返回特征解读

第二讲 分式

exams/lecture-02-rational-expressions/第二讲_分式.pdf · HS-MATH-1024-v2.1-solution-aware

12 题13 个小问/题组
1

例题解析 · 分式/绝对值/分类讨论

化简：1−x−x2+x31−2∣x∣+x2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{3}}}{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{|}\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{|}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{x}^{\htmlData{tutor-start=33,tutor-end=34}{2}}}。

答案：x>=0 时为 x+1；x<0 时为 (1-x)^2/(1+x)，且 x!=±1

题目标签：含绝对值分母的分式化简

知识点
解题操作与技能

解题过程

按绝对值符号分类化简

在原分式有定义的前提下写出分段结果

(1)
分解分子并整理分母

分子分组后可连续提出 1−x\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}；分母利用 x2=∣x∣2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{|}^{\htmlData{tutor-start=11,tutor-end=12}{2}} 写成完全平方。原式有定义还要求 ∣x∣≠1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\ne\htmlData{tutor-start=6,tutor-end=7}{1}。

详细展开：首先处理分子 1−x−x2+x3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{3}}。我们采用分组分解法，将前两项 1−x\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x} 与后两项 −x2+x3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}} 分别组合。观察后一组，提取公因式 −x2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}，得到 −x2(1−x)\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}。此时整个分子变为 (1−x)−x2(1−x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{)}，两项均含有公因式 (1−x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}，提取后得到 (1−x)(1−x2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}。接着利用平方差公式 a2−b2=(a−b)(a+b)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{a}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{)}，将 1−x2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}} 分解为 (1−x)(1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}，因此分子最终化为 (1−x)2(1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}。对于分母 1−2∣x∣+x2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}，注意到对任意实数 x\htmlData{tutor-start=0,tutor-end=1}{x} 都有 x2=∣x∣2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{|}^{\htmlData{tutor-start=11,tutor-end=12}{2}}，因此分母可写为 1−2∣x∣+∣x∣2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{|}^{\htmlData{tutor-start=12,tutor-end=13}{2}}。这符合完全平方公式 a2−2ab+b2=(a−b)2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{b}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}} 的结构，其中 a=1,b=∣x∣\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{|}，故分母化为 (1−∣x∣)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}}。最后，根据分式有意义的条件，分母不能为零，即 (1−∣x∣)2≠0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \neq \htmlData{tutor-start=17,tutor-end=18}{0}，解得 ∣x∣≠1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|} \neq \htmlData{tutor-start=9,tutor-end=10}{1}，这意味着 x≠1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1} 且 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。这些限制条件在后续化简中必须始终保留。

1−x−x2+x3=(1−x)+(−x2+x3)=(1−x)−x2(1−x)=(1−x)(1−x2)=(1−x)(1−x)(1+x)=(1−x)2(1+x),1−2∣x∣+x2=1−2∣x∣+∣x∣2=(1−∣x∣)2.\begin{aligned}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{x}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{3}}&\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{x}^{\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{x}^{\htmlData{tutor-start=49,tutor-end=50}{3}}\htmlData{tutor-start=51,tutor-end=52}{)}\\&\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{x}\htmlData{tutor-start=60,tutor-end=61}{)}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{x}^{\htmlData{tutor-start=65,tutor-end=66}{2}}\htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{1}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{x}\htmlData{tutor-start=71,tutor-end=72}{)}\\&\htmlData{tutor-start=75,tutor-end=76}{=}\htmlData{tutor-start=76,tutor-end=77}{(}\htmlData{tutor-start=77,tutor-end=78}{1}\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=80}{x}\htmlData{tutor-start=80,tutor-end=81}{)}\htmlData{tutor-start=81,tutor-end=82}{(}\htmlData{tutor-start=82,tutor-end=83}{1}\htmlData{tutor-start=83,tutor-end=84}{-}\htmlData{tutor-start=84,tutor-end=85}{x}^{\htmlData{tutor-start=87,tutor-end=88}{2}}\htmlData{tutor-start=89,tutor-end=90}{)}\\&\htmlData{tutor-start=93,tutor-end=94}{=}\htmlData{tutor-start=94,tutor-end=95}{(}\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=97}{-}\htmlData{tutor-start=97,tutor-end=98}{x}\htmlData{tutor-start=98,tutor-end=99}{)}\htmlData{tutor-start=99,tutor-end=100}{(}\htmlData{tutor-start=100,tutor-end=101}{1}\htmlData{tutor-start=101,tutor-end=102}{-}\htmlData{tutor-start=102,tutor-end=103}{x}\htmlData{tutor-start=103,tutor-end=104}{)}\htmlData{tutor-start=104,tutor-end=105}{(}\htmlData{tutor-start=105,tutor-end=106}{1}\htmlData{tutor-start=106,tutor-end=107}{+}\htmlData{tutor-start=107,tutor-end=108}{x}\htmlData{tutor-start=108,tutor-end=109}{)}\\&\htmlData{tutor-start=112,tutor-end=113}{=}\htmlData{tutor-start=113,tutor-end=114}{(}\htmlData{tutor-start=114,tutor-end=115}{1}\htmlData{tutor-start=115,tutor-end=116}{-}\htmlData{tutor-start=116,tutor-end=117}{x}\htmlData{tutor-start=117,tutor-end=118}{)}^{\htmlData{tutor-start=120,tutor-end=121}{2}}\htmlData{tutor-start=122,tutor-end=123}{(}\htmlData{tutor-start=123,tutor-end=124}{1}\htmlData{tutor-start=124,tutor-end=125}{+}\htmlData{tutor-start=125,tutor-end=126}{x}\htmlData{tutor-start=126,tutor-end=127}{)}\htmlData{tutor-start=127,tutor-end=128}{,}\\\htmlData{tutor-start=130,tutor-end=131}{1}\htmlData{tutor-start=131,tutor-end=132}{-}\htmlData{tutor-start=132,tutor-end=133}{2}\htmlData{tutor-start=133,tutor-end=134}{|}\htmlData{tutor-start=134,tutor-end=135}{x}\htmlData{tutor-start=135,tutor-end=136}{|}\htmlData{tutor-start=136,tutor-end=137}{+}\htmlData{tutor-start=137,tutor-end=138}{x}^{\htmlData{tutor-start=140,tutor-end=141}{2}}&\htmlData{tutor-start=143,tutor-end=144}{=}\htmlData{tutor-start=144,tutor-end=145}{1}\htmlData{tutor-start=145,tutor-end=146}{-}\htmlData{tutor-start=146,tutor-end=147}{2}\htmlData{tutor-start=147,tutor-end=148}{|}\htmlData{tutor-start=148,tutor-end=149}{x}\htmlData{tutor-start=149,tutor-end=150}{|}\htmlData{tutor-start=150,tutor-end=151}{+}\htmlData{tutor-start=151,tutor-end=152}{|}\htmlData{tutor-start=152,tutor-end=153}{x}\htmlData{tutor-start=153,tutor-end=154}{|}^{\htmlData{tutor-start=156,tutor-end=157}{2}}\\&\htmlData{tutor-start=161,tutor-end=162}{=}\htmlData{tutor-start=162,tutor-end=163}{(}\htmlData{tutor-start=163,tutor-end=164}{1}\htmlData{tutor-start=164,tutor-end=165}{-}\htmlData{tutor-start=165,tutor-end=166}{|}\htmlData{tutor-start=166,tutor-end=167}{x}\htmlData{tutor-start=167,tutor-end=168}{|}\htmlData{tutor-start=168,tutor-end=169}{)}^{\htmlData{tutor-start=171,tutor-end=172}{2}}\htmlData{tutor-start=173,tutor-end=174}{.}\end{aligned}
(2)
处理 x≥0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0} 的情形

此时 ∣x∣=x\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{x}，分母为 (1−x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。约去非零因子后得到 x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}，并保留 x≠1\htmlData{tutor-start=0,tutor-end=1}{x}\ne\htmlData{tutor-start=4,tutor-end=5}{1}。

详细展开：接下来讨论 x≥0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{0} 的情形。根据绝对值的定义，当 x≥0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{0} 时，∣x∣=x\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{x}。将此代入第一步得到的分母形式 (1−∣x∣)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}}，分母变为 (1−x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。此时原分式化为 (1−x)2(1+x)(1−x)2\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}。由于我们在第一步中已经确定了定义域要求 x≠1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}，因此在 x≥0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{0} 且 x≠1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1} 的范围内，因子 (1−x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} 恒不为零。我们可以安全地约去分子和分母中的公因式 (1−x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。约分后，分子剩余 1+x\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{x}，分母剩余 1\htmlData{tutor-start=0,tutor-end=1}{1}。因此，在该分支下，原式化简为 x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}。需要注意的是，虽然化简结果 x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1} 在 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 处有定义，但原分式在 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 处无定义，因此必须明确标注条件 x≠1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}。综上，当 x≥0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{0} 且 x≠1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1} 时，结果为 x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}。

x≥0, x≠1:(1−x)2(1+x)(1−x)2=x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=8}{\ }\htmlData{tutor-start=8,tutor-end=9}{x}\ne\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{:}\quad \frac{\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=36}{^{2}(}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=40}{)}}{\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{x}\htmlData{tutor-start=46,tutor-end=47}{)}^{\htmlData{tutor-start=49,tutor-end=50}{2}}}\htmlData{tutor-start=52,tutor-end=53}{=}\htmlData{tutor-start=53,tutor-end=54}{x}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{1}
(3)
处理 x<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0} 的情形

此时 ∣x∣=−x\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}，分母为 (1+x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。只能约去一个 1+x\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{x}，且 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x}\ne\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}。

详细展开：最后讨论 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 的情形。根据绝对值的定义，当 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 时，∣x∣=−x\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}。将此代入第一步得到的分母形式 (1−∣x∣)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}}，分母变为 (1−(−x))2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{)}^{\htmlData{tutor-start=10,tutor-end=11}{2}}，即 (1+x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。此时原分式化为 (1−x)2(1+x)(1+x)2\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}。同样，依据第一步确定的定义域，我们知道 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。在 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 且 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} 的范围内，因子 (1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} 不为零，因此 (1+x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} 也不为零。我们可以约去分子和分母中的一个公因式 (1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}。注意，分子中 (1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} 的次数是 1，分母中是 2，约分后分母仍剩余一个 (1+x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}。分子剩余 (1−x)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。因此，在该分支下，原式化简为 (1−x)21+x\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}}。必须再次强调，尽管化简后的表达式在 x=−1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 处看似可以通过极限理解，但原分式在 x=−1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 处分母为零，无定义，因此必须保留条件 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}。综上，当 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 且 x≠−1\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} 时，结果为 (1−x)21+x\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}}。

x<0, x≠−1:(1−x)2(1+x)(1+x)2=(1−x)21+xx<0,\ x\ne-1:\quad \frac{(1-x)^{2}(1+x)}{(1+x)^{2}}=\frac{(1-x)^{2}}{1+x}
相似题目从题目升维到题型

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2

例题解析 · 分式/因式分解/通分

化简：aa3+a2b+ab2+b3+ba3−a2b+ab2−b3+1a2−b2−1a2+b2−a2+3b2a4−b4\frac{a}{a^{3}+a^{2}b+ab^{2}+b^{3}}+\frac{b}{a^{3}-a^{2}b+ab^{2}-b^{3}}+\frac1{a^{2}-b^{2}}-\frac1{a^{2}+b^{2}}-\frac{a^{2}+3b^{2}}{a^{4}-b^{4}}。

答案：0\htmlData{tutor-start=0,tutor-end=1}{0}

题目标签：多项式因式分解后通分

知识点
解题操作与技能

解题过程

分解分母后逐组通分

证明整个分式和为 0

(1)
分解四个多项式分母

先识别立方分组和平方差，使所有分母都由 a−b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}、a+b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}、a2+b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} 组成。

详细展开：观察原式中的四个分母，发现它们都是关于 a\htmlData{tutor-start=0,tutor-end=1}{a} 和 b\htmlData{tutor-start=0,tutor-end=1}{b} 的多项式。为了后续通分，我们需要先对每个分母进行因式分解。首先看第一个分母 a3+a2b+ab2+b3\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{3}}，我们可以采用分组分解法：前两项提取公因式 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，后两项提取公因式 b2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，得到 a2(a+b)+b2(a+b)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{)}，再提取公因式 (a+b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}，最终分解为 (a+b)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{)}。接着看第二个分母 a3−a2b+ab2−b3\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{3}}，同样分组：a2(a−b)+b2(a−b)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{)}，提取 (a−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)} 后得到 (a−b)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{)}。第三个分母 a2−b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} 是标准的平方差公式，直接分解为 (a−b)(a+b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}。第四个分母 a4−b4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{4}} 可以看作 (a2)2−(b2)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{b}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}}，先分解为 (a2−b2)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}，再将 a2−b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} 继续分解，得到 (a−b)(a+b)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{b}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{)}。通过这一步，所有分母都转化为了基本因式 a−b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}、a+b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b} 和 a2+b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} 的乘积形式。

a3+a2b+ab2+b3=(a+b)(a2+b2),a3−a2b+ab2−b3=(a−b)(a2+b2),a4−b4=(a2−b2)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{b}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{,}\quad \htmlData{tutor-start=51,tutor-end=52}{a}^{\htmlData{tutor-start=54,tutor-end=55}{3}}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{a}^{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{b}\htmlData{tutor-start=63,tutor-end=64}{+}\htmlData{tutor-start=64,tutor-end=65}{a}\htmlData{tutor-start=65,tutor-end=66}{b}^{\htmlData{tutor-start=68,tutor-end=69}{2}}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{b}^{\htmlData{tutor-start=74,tutor-end=75}{3}}\htmlData{tutor-start=76,tutor-end=77}{=}\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{a}\htmlData{tutor-start=79,tutor-end=80}{-}\htmlData{tutor-start=80,tutor-end=81}{b}\htmlData{tutor-start=81,tutor-end=82}{)}\htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{a}^{\htmlData{tutor-start=86,tutor-end=87}{2}}\htmlData{tutor-start=88,tutor-end=89}{+}\htmlData{tutor-start=89,tutor-end=90}{b}^{\htmlData{tutor-start=92,tutor-end=93}{2}}\htmlData{tutor-start=94,tutor-end=95}{)}\htmlData{tutor-start=95,tutor-end=96}{,}\quad \htmlData{tutor-start=102,tutor-end=103}{a}^{\htmlData{tutor-start=105,tutor-end=106}{4}}\htmlData{tutor-start=107,tutor-end=108}{-}\htmlData{tutor-start=108,tutor-end=109}{b}^{\htmlData{tutor-start=111,tutor-end=112}{4}}\htmlData{tutor-start=113,tutor-end=114}{=}\htmlData{tutor-start=114,tutor-end=115}{(}\htmlData{tutor-start=115,tutor-end=116}{a}^{\htmlData{tutor-start=118,tutor-end=119}{2}}\htmlData{tutor-start=120,tutor-end=121}{-}\htmlData{tutor-start=121,tutor-end=122}{b}^{\htmlData{tutor-start=124,tutor-end=125}{2}}\htmlData{tutor-start=126,tutor-end=127}{)}\htmlData{tutor-start=127,tutor-end=128}{(}\htmlData{tutor-start=128,tutor-end=129}{a}^{\htmlData{tutor-start=131,tutor-end=132}{2}}\htmlData{tutor-start=133,tutor-end=134}{+}\htmlData{tutor-start=134,tutor-end=135}{b}^{\htmlData{tutor-start=137,tutor-end=138}{2}}\htmlData{tutor-start=139,tutor-end=140}{)}
(2)
合并前两个分式

前两个分式通分后，分子中的交叉项抵消。

详细展开：根据上一步的分解结果，前两个分式分别为 a(a+b)(a2+b2)\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{b}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}} 和 b(a−b)(a2+b2)\frac{\htmlData{tutor-start=6,tutor-end=7}{b}}{\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{b}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}}。观察发现，这两个分式的分母都含有公因式 (a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}，而另一个因式分别是 (a+b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)} 和 (a−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}。因此，它们的最小公分母是 (a+b)(a−b)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{b}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{)}，即 (a2−b2)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}。我们将这两个分式通分：第一个分式分子分母同乘 (a−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}，变为 a(a−b)(a2−b2)(a2+b2)\frac{\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{)}}{\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{b}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{a}^{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{b}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{)}}；第二个分式分子分母同乘 (a+b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}，变为 b(a+b)(a2−b2)(a2+b2)\frac{\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{)}}{\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{b}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{a}^{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{b}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{)}}。将分子相加：a(a−b)+b(a+b)=a2−ab+ab+b2=a2+b2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{b}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{b}^{\htmlData{tutor-start=31,tutor-end=32}{2}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{a}^{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{b}^{\htmlData{tutor-start=45,tutor-end=46}{2}}。此时，分式变为 a2+b2(a2−b2)(a2+b2)\frac{\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}{\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{b}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{a}^{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{b}^{\htmlData{tutor-start=42,tutor-end=43}{2}}\htmlData{tutor-start=44,tutor-end=45}{)}}。由于 a2+b2≠0\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \neq \htmlData{tutor-start=17,tutor-end=18}{0}（在实数范围内，除非 a=b=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0}，但此时原式无意义），我们可以约去分子和分母中的 (a2+b2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}，最终得到 1a2−b2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}。这一步成功将前两项合并为一个简单的分式。

a(a+b)(a2+b2)+b(a−b)(a2+b2)=a(a−b)+b(a+b)(a2−b2)(a2+b2)=1a2−b2\frac{a}{(a+b)(a^{2}+b^{2})}+\frac{b}{(a-b)(a^{2}+b^{2})}=\frac{a(a-b)+b(a+b)}{(a^{2}-b^{2})(a^{2}+b^{2})}=\frac1{a^{2}-b^{2}}
(3)
统一到 a4−b4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{4}} 并化简

把剩余四项通分，检查分子后恰好为 0。原式要求各分母非零。

详细展开：经过前两步的处理，原式现在简化为：1a2−b2+1a2−b2−1a2+b2−a2+3b2a4−b4\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}} \htmlData{tutor-start=22,tutor-end=23}{+} \frac{\htmlData{tutor-start=30,tutor-end=31}{1}}{\htmlData{tutor-start=33,tutor-end=34}{a}^{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{b}^{\htmlData{tutor-start=42,tutor-end=43}{2}}} \htmlData{tutor-start=46,tutor-end=47}{-} \frac{\htmlData{tutor-start=54,tutor-end=55}{1}}{\htmlData{tutor-start=57,tutor-end=58}{a}^{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{b}^{\htmlData{tutor-start=66,tutor-end=67}{2}}} \htmlData{tutor-start=70,tutor-end=71}{-} \frac{\htmlData{tutor-start=78,tutor-end=79}{a}^{\htmlData{tutor-start=81,tutor-end=82}{2}}\htmlData{tutor-start=83,tutor-end=84}{+}\htmlData{tutor-start=84,tutor-end=85}{3}\htmlData{tutor-start=85,tutor-end=86}{b}^{\htmlData{tutor-start=88,tutor-end=89}{2}}}{\htmlData{tutor-start=92,tutor-end=93}{a}^{\htmlData{tutor-start=95,tutor-end=96}{4}}\htmlData{tutor-start=97,tutor-end=98}{-}\htmlData{tutor-start=98,tutor-end=99}{b}^{\htmlData{tutor-start=101,tutor-end=102}{4}}}。注意，原题中的第三项是 +1a2−b2\htmlData{tutor-start=0,tutor-end=1}{+}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}，加上我们刚刚合并前两项得到的 1a2−b2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}，现在共有两个 1a2−b2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}，合并后为 2a2−b2\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}。剩下的项是 −1a2+b2\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} 和 −a2+3b2a4−b4\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}{\htmlData{tutor-start=21,tutor-end=22}{a}^{\htmlData{tutor-start=24,tutor-end=25}{4}}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{b}^{\htmlData{tutor-start=30,tutor-end=31}{4}}}。现在我们需要将这三部分通分。观察分母：a2−b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}、a2+b2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} 和 a4−b4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{4}}。因为 a4−b4=(a2−b2)(a2+b2)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{4}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{b}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{a}^{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{b}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{)}，所以最小公分母就是 a4−b4\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{4}}。我们将前两项通分：2a2−b2=2(a2+b2)a4−b4\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}} \htmlData{tutor-start=22,tutor-end=23}{=} \frac{\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{)}}{\htmlData{tutor-start=46,tutor-end=47}{a}^{\htmlData{tutor-start=49,tutor-end=50}{4}}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{b}^{\htmlData{tutor-start=55,tutor-end=56}{4}}}，−1a2+b2=−a2−b2a4−b4\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{-}\frac{\htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}}{\htmlData{tutor-start=45,tutor-end=46}{a}^{\htmlData{tutor-start=48,tutor-end=49}{4}}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{b}^{\htmlData{tutor-start=54,tutor-end=55}{4}}}。最后一项保持不变。现在将所有分子合并：2(a2+b2)−(a2−b2)−(a2+3b2)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{a}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{b}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{)} \htmlData{tutor-start=31,tutor-end=32}{-} \htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{a}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{3}\htmlData{tutor-start=41,tutor-end=42}{b}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{)}。展开括号：2a2+2b2−a2+b2−a2−3b2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{b}^{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{a}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{b}^{\htmlData{tutor-start=40,tutor-end=41}{2}}。合并同类项：a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 的系数为 2−1−1=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，b2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 的系数为 2+1−3=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}。因此分子为 0，整个分式的值为 0。需注意定义域要求 a≠±b\htmlData{tutor-start=0,tutor-end=1}{a} \neq \htmlData{tutor-start=7,tutor-end=11}{\pm }\htmlData{tutor-start=11,tutor-end=12}{b} 且 a2+b2≠0\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \neq \htmlData{tutor-start=17,tutor-end=18}{0}。

2a2−b2−1a2+b2−a2+3b2a4−b4=2(a2+b2)−(a2−b2)−(a2+3b2)a4−b4=0\frac{2}{a^{2}-b^{2}}-\frac1{a^{2}+b^{2}}-\frac{a^{2}+3b^{2}}{a^{4}-b^{4}}=\frac{2(a^{2}+b^{2})-(a^{2}-b^{2})-(a^{2}+3b^{2})}{a^{4}-b^{4}}=0
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3

例题解析 · 分式/换元/倒数关系

计算下列两个分式： （1）n2m2+m2n2+2n3m3−m3n3−3(nm−mn)\frac{\frac{\htmlData{tutor-start=12,tutor-end=13}{n}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}{\htmlData{tutor-start=19,tutor-end=20}{m}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}\htmlData{tutor-start=25,tutor-end=26}{+}\frac{\htmlData{tutor-start=32,tutor-end=33}{m}^{\htmlData{tutor-start=35,tutor-end=36}{2}}}{\htmlData{tutor-start=39,tutor-end=40}{n}^{\htmlData{tutor-start=42,tutor-end=43}{2}}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{2}}{\frac{\htmlData{tutor-start=55,tutor-end=56}{n}^{\htmlData{tutor-start=58,tutor-end=59}{3}}}{\htmlData{tutor-start=62,tutor-end=63}{m}^{\htmlData{tutor-start=65,tutor-end=66}{3}}}\htmlData{tutor-start=68,tutor-end=69}{-}\frac{\htmlData{tutor-start=75,tutor-end=76}{m}^{\htmlData{tutor-start=78,tutor-end=79}{3}}}{\htmlData{tutor-start=82,tutor-end=83}{n}^{\htmlData{tutor-start=85,tutor-end=86}{3}}}\htmlData{tutor-start=88,tutor-end=89}{-}\htmlData{tutor-start=89,tutor-end=90}{3}\htmlData{tutor-start=90,tutor-end=91}{(}\frac{\htmlData{tutor-start=97,tutor-end=98}{n}}{\htmlData{tutor-start=100,tutor-end=101}{m}}\htmlData{tutor-start=102,tutor-end=103}{-}\frac{\htmlData{tutor-start=109,tutor-end=110}{m}}{\htmlData{tutor-start=112,tutor-end=113}{n}}\htmlData{tutor-start=114,tutor-end=115}{)}}； （2）nm+mnn2m2+m2n2−2\frac{\frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{m}}\htmlData{tutor-start=17,tutor-end=18}{+}\frac{\htmlData{tutor-start=24,tutor-end=25}{m}}{\htmlData{tutor-start=27,tutor-end=28}{n}}}{\frac{\htmlData{tutor-start=37,tutor-end=38}{n}^{\htmlData{tutor-start=40,tutor-end=41}{2}}}{\htmlData{tutor-start=44,tutor-end=45}{m}^{\htmlData{tutor-start=47,tutor-end=48}{2}}}\htmlData{tutor-start=50,tutor-end=51}{+}\frac{\htmlData{tutor-start=57,tutor-end=58}{m}^{\htmlData{tutor-start=60,tutor-end=61}{2}}}{\htmlData{tutor-start=64,tutor-end=65}{n}^{\htmlData{tutor-start=67,tutor-end=68}{2}}}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{2}}。

答案：见两个小问

题目标签：用倒数换元化简繁分式

知识点
解题操作与技能

解题过程

（1）第（1）个繁分式

用倒数换元化简第一个表达式

(1)
引入互为倒数的整体

令 u=nm\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}、v=mn\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{m}}{\htmlData{tutor-start=11,tutor-end=12}{n}}，则 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。分子立即成为 (u+v)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。

详细展开：观察原式分子 n2m2+m2n2+2\frac{\htmlData{tutor-start=6,tutor-end=7}{n}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{+}\frac{\htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{n}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{2}，发现其中反复出现互为倒数的项 nm\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{m}} 与 mn\frac{\htmlData{tutor-start=6,tutor-end=7}{m}}{\htmlData{tutor-start=9,tutor-end=10}{n}}。为了简化书写并利用代数恒等式，我们引入换元：令 u=nm\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}，则其倒数 v=mn=1u\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{m}}{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{=}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{u}}，因此满足关键约束条件 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。此时，原分子中的平方项可写为 u2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 和 v2\htmlData{tutor-start=0,tutor-end=1}{v}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。根据完全平方公式 (a+b)2=a2+2ab+b2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{2}}，我们将 u2+v2+2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2} 进行变形。由于 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，中间项 2\htmlData{tutor-start=0,tutor-end=1}{2} 恰好等于 2uv\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{v}。因此，分子 u2+v2+2=u2+2uv+v2=(u+v)2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{u}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{u}\htmlData{tutor-start=24,tutor-end=25}{v}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{v}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{u}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{v}\htmlData{tutor-start=38,tutor-end=39}{)}^{\htmlData{tutor-start=41,tutor-end=42}{2}}。这一步将复杂的分式加法转化为了一个简洁的平方形式，为后续约分打下基础。

u=nm,v=mn,uv=1,u2+v2+2=(u+v)2\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}\htmlData{tutor-start=13,tutor-end=14}{,}\quad \htmlData{tutor-start=20,tutor-end=21}{v}\htmlData{tutor-start=21,tutor-end=22}{=}\frac{\htmlData{tutor-start=28,tutor-end=29}{m}}{\htmlData{tutor-start=31,tutor-end=32}{n}}\htmlData{tutor-start=33,tutor-end=34}{,}\quad \htmlData{tutor-start=40,tutor-end=41}{u}\htmlData{tutor-start=41,tutor-end=42}{v}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{,}\quad \htmlData{tutor-start=51,tutor-end=52}{u}^{\htmlData{tutor-start=54,tutor-end=55}{2}}\htmlData{tutor-start=56,tutor-end=57}{+}\htmlData{tutor-start=57,tutor-end=58}{v}^{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{2}\htmlData{tutor-start=64,tutor-end=65}{=}\htmlData{tutor-start=65,tutor-end=66}{(}\htmlData{tutor-start=66,tutor-end=67}{u}\htmlData{tutor-start=67,tutor-end=68}{+}\htmlData{tutor-start=68,tutor-end=69}{v}\htmlData{tutor-start=69,tutor-end=70}{)}^{\htmlData{tutor-start=72,tutor-end=73}{2}}
(2)
把分母化成立方

利用 u3−v3=(u−v)(u2+uv+v2)\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{v}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{u}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{u}\htmlData{tutor-start=25,tutor-end=26}{v}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{v}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{)} 和 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。

详细展开：接下来处理原式的分母部分：n3m3−m3n3−3(nm−mn)\frac{\htmlData{tutor-start=6,tutor-end=7}{n}^{\htmlData{tutor-start=9,tutor-end=10}{3}}}{\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{3}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{3}}}{\htmlData{tutor-start=33,tutor-end=34}{n}^{\htmlData{tutor-start=36,tutor-end=37}{3}}}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{3}\htmlData{tutor-start=41,tutor-end=42}{(}\frac{\htmlData{tutor-start=48,tutor-end=49}{n}}{\htmlData{tutor-start=51,tutor-end=52}{m}}\htmlData{tutor-start=53,tutor-end=54}{-}\frac{\htmlData{tutor-start=60,tutor-end=61}{m}}{\htmlData{tutor-start=63,tutor-end=64}{n}}\htmlData{tutor-start=65,tutor-end=66}{)}。利用之前的换元 u=nm,v=mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{v}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\htmlData{tutor-start=23,tutor-end=24}{m}}{\htmlData{tutor-start=26,tutor-end=27}{n}}，分母变为 u3−v3−3(u−v)\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{u}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{v}\htmlData{tutor-start=17,tutor-end=18}{)}。我们需要利用立方差公式 a3−b3=(a−b)(a2+ab+b2)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{a}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{a}\htmlData{tutor-start=25,tutor-end=26}{b}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{b}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{)} 对前两项进行分解。于是 u3−v3=(u−v)(u2+uv+v2)\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{3}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{u}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{v}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{u}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{u}\htmlData{tutor-start=27,tutor-end=28}{v}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{v}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{)}。将此代入分母表达式，提取公因式 (u−v)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}，得到：(u−v)[(u2+uv+v2)−3]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{[}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{u}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{v}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{]}。利用已知条件 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，括号内变为 u2+1+v2−3=u2+v2−2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{3} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{u}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{v}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{2}。再次观察，u2+v2−2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2} 恰好符合完全平方公式的逆运算，即 u2−2uv+v2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{v}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{v}^{\htmlData{tutor-start=13,tutor-end=14}{2}}（因为 2uv=2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}），所以 u2+v2−2=(u−v)2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{u}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{v}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}}。最终，分母化为 (u−v)⋅(u−v)2=(u−v)3\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=12}{\cdot }\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{v}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{u}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{v}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{3}}。

u3−v3−3(u−v)=(u−v)(u2+v2−2)=(u−v)3\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{u}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{v}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{u}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{v}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{u}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{v}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{)}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{u}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{v}\htmlData{tutor-start=44,tutor-end=45}{)}^{\htmlData{tutor-start=47,tutor-end=48}{3}}
(3)
还原为 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n}

代回 u+v=m2+n2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\frac{\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{n}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}{\htmlData{tutor-start=23,tutor-end=24}{m}\htmlData{tutor-start=24,tutor-end=25}{n}}、u−v=n2−m2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\frac{\htmlData{tutor-start=10,tutor-end=11}{n}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{m}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}{\htmlData{tutor-start=23,tutor-end=24}{m}\htmlData{tutor-start=24,tutor-end=25}{n}}。

详细展开：现在原式已化简为 (u+v)2(u−v)3\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{v}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{u}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{v}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{3}}}。最后一步是将 u,v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{v} 还原为 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{n}。首先计算 u+v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v} 和 u−v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v} 的通分形式：u+v=nm+mn=n2+m2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{m}} \htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=27}{m}}{\htmlData{tutor-start=29,tutor-end=30}{n}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{n}^{\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{m}^{\htmlData{tutor-start=49,tutor-end=50}{2}}}{\htmlData{tutor-start=53,tutor-end=54}{m}\htmlData{tutor-start=54,tutor-end=55}{n}}；u−v=nm−mn=n2−m2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{m}} \htmlData{tutor-start=18,tutor-end=19}{-} \frac{\htmlData{tutor-start=26,tutor-end=27}{m}}{\htmlData{tutor-start=29,tutor-end=30}{n}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{n}^{\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{m}^{\htmlData{tutor-start=49,tutor-end=50}{2}}}{\htmlData{tutor-start=53,tutor-end=54}{m}\htmlData{tutor-start=54,tutor-end=55}{n}}。将这两个结果代入化简后的式子：分子为 (m2+n2mn)2=(m2+n2)2m2n2\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{m}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{n}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{n}}\htmlData{tutor-start=23,tutor-end=24}{)}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{m}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{n}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{)}^{\htmlData{tutor-start=52,tutor-end=53}{2}}}{\htmlData{tutor-start=56,tutor-end=57}{m}^{\htmlData{tutor-start=59,tutor-end=60}{2}}\htmlData{tutor-start=61,tutor-end=62}{n}^{\htmlData{tutor-start=64,tutor-end=65}{2}}}；分母为 (n2−m2mn)3=(n2−m2)3m3n3\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{n}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{n}}\htmlData{tutor-start=23,tutor-end=24}{)}^{\htmlData{tutor-start=26,tutor-end=27}{3}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{n}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{m}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{)}^{\htmlData{tutor-start=52,tutor-end=53}{3}}}{\htmlData{tutor-start=56,tutor-end=57}{m}^{\htmlData{tutor-start=59,tutor-end=60}{3}}\htmlData{tutor-start=61,tutor-end=62}{n}^{\htmlData{tutor-start=64,tutor-end=65}{3}}}。执行除法运算，即分子乘以分母的倒数：(m2+n2)2m2n2⋅m3n3(n2−m2)3\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{m}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{n}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{m}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{n}^{\htmlData{tutor-start=33,tutor-end=34}{2}}} \htmlData{tutor-start=37,tutor-end=43}{\cdot }\frac{\htmlData{tutor-start=49,tutor-end=50}{m}^{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{n}^{\htmlData{tutor-start=57,tutor-end=58}{3}}}{\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{n}^{\htmlData{tutor-start=65,tutor-end=66}{2}}\htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{m}^{\htmlData{tutor-start=71,tutor-end=72}{2}}\htmlData{tutor-start=73,tutor-end=74}{)}^{\htmlData{tutor-start=76,tutor-end=77}{3}}}。约去公共因子 m2n2\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{n}^{\htmlData{tutor-start=8,tutor-end=9}{2}}，分子剩余 mn\htmlData{tutor-start=0,tutor-end=1}{m} \htmlData{tutor-start=2,tutor-end=3}{n}，分母剩余 (n2−m2)3\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{m}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{3}}。最终结果为 mn(m2+n2)2(n2−m2)3\frac{\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{m}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{n}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}}}{\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{n}^{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{m}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{)}^{\htmlData{tutor-start=42,tutor-end=43}{3}}}。注意需保证 m,n≠0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n} \neq \htmlData{tutor-start=9,tutor-end=10}{0} 且 n2≠m2\htmlData{tutor-start=0,tutor-end=1}{n}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \neq \htmlData{tutor-start=11,tutor-end=12}{m}^{\htmlData{tutor-start=14,tutor-end=15}{2}}。

(u+v)2(u−v)3=mn(m2+n2)2(n2−m2)3\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{v}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{u}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{v}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{3}}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{m}\htmlData{tutor-start=35,tutor-end=36}{n}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{m}^{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{n}^{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{)}^{\htmlData{tutor-start=51,tutor-end=52}{2}}}{\htmlData{tutor-start=55,tutor-end=56}{(}\htmlData{tutor-start=56,tutor-end=57}{n}^{\htmlData{tutor-start=59,tutor-end=60}{2}}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{m}^{\htmlData{tutor-start=65,tutor-end=66}{2}}\htmlData{tutor-start=67,tutor-end=68}{)}^{\htmlData{tutor-start=70,tutor-end=71}{3}}}

（2）第（2）个繁分式

用同一换元化简第二个表达式

(1)
沿用倒数换元

仍令 u=nm\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}、v=mn\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{m}}{\htmlData{tutor-start=11,tutor-end=12}{n}}，且 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。分母正好是 (u−v)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。

详细展开：对于第二个分式 nm+mnn2m2+m2n2−2\frac{\frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{m}}\htmlData{tutor-start=17,tutor-end=18}{+}\frac{\htmlData{tutor-start=24,tutor-end=25}{m}}{\htmlData{tutor-start=27,tutor-end=28}{n}}}{\frac{\htmlData{tutor-start=37,tutor-end=38}{n}^{\htmlData{tutor-start=40,tutor-end=41}{2}}}{\htmlData{tutor-start=44,tutor-end=45}{m}^{\htmlData{tutor-start=47,tutor-end=48}{2}}}\htmlData{tutor-start=50,tutor-end=51}{+}\frac{\htmlData{tutor-start=57,tutor-end=58}{m}^{\htmlData{tutor-start=60,tutor-end=61}{2}}}{\htmlData{tutor-start=64,tutor-end=65}{n}^{\htmlData{tutor-start=67,tutor-end=68}{2}}}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{2}}，我们沿用相同的换元策略。令 u=nm,v=mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{n}}{\htmlData{tutor-start=11,tutor-end=12}{m}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{v}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\htmlData{tutor-start=23,tutor-end=24}{m}}{\htmlData{tutor-start=26,tutor-end=27}{n}}，则 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。分子直接变为 u+v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v}。重点分析分母：n2m2+m2n2−2\frac{\htmlData{tutor-start=6,tutor-end=7}{n}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{+}\frac{\htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{n}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{2} 对应 u2+v2−2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}。利用 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，我们将常数 −2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2} 写为 −2uv\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{u}\htmlData{tutor-start=3,tutor-end=4}{v}。于是分母变为 u2−2uv+v2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{v}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{v}^{\htmlData{tutor-start=13,tutor-end=14}{2}}。根据完全平方公式 (a−b)2=a2−2ab+b2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{2}}，这正是 (u−v)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。因此，原分式在换元后简化为 u+v(u−v)2\frac{\htmlData{tutor-start=6,tutor-end=7}{u}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}}{\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{u}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}。这一步展示了同一组变量在不同幂次下的结构一致性。

u2+v2−2uv=(u−v)2\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{u}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{v}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}}
(2)
代回原变量

把和与差分别改写为 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n} 的分式，再约去公共因子。

详细展开：最后，将 u+v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v} 和 u−v\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v} 还原为 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n} 的表达式。由前可知，u+v=m2+n2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{m}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{n}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{m}\htmlData{tutor-start=26,tutor-end=27}{n}}，u−v=n2−m2mn\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{v} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{n}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{m}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{m}\htmlData{tutor-start=26,tutor-end=27}{n}}。代入简化后的分式 u+v(u−v)2\frac{\htmlData{tutor-start=6,tutor-end=7}{u}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}}{\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{u}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}：分子为 m2+n2mn\frac{\htmlData{tutor-start=6,tutor-end=7}{m}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{n}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}{\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{n}}；分母为 (n2−m2mn)2=(n2−m2)2m2n2\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{n}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{n}}\htmlData{tutor-start=23,tutor-end=24}{)}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{n}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{m}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{)}^{\htmlData{tutor-start=52,tutor-end=53}{2}}}{\htmlData{tutor-start=56,tutor-end=57}{m}^{\htmlData{tutor-start=59,tutor-end=60}{2}}\htmlData{tutor-start=61,tutor-end=62}{n}^{\htmlData{tutor-start=64,tutor-end=65}{2}}}。执行除法：m2+n2mn÷(n2−m2)2m2n2=m2+n2mn⋅m2n2(n2−m2)2\frac{\htmlData{tutor-start=6,tutor-end=7}{m}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{n}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}{\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{n}} \htmlData{tutor-start=23,tutor-end=28}{\div }\frac{\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{n}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{m}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{)}^{\htmlData{tutor-start=49,tutor-end=50}{2}}}{\htmlData{tutor-start=53,tutor-end=54}{m}^{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=59}{n}^{\htmlData{tutor-start=61,tutor-end=62}{2}}} \htmlData{tutor-start=65,tutor-end=66}{=} \frac{\htmlData{tutor-start=73,tutor-end=74}{m}^{\htmlData{tutor-start=76,tutor-end=77}{2}}\htmlData{tutor-start=78,tutor-end=79}{+}\htmlData{tutor-start=79,tutor-end=80}{n}^{\htmlData{tutor-start=82,tutor-end=83}{2}}}{\htmlData{tutor-start=86,tutor-end=87}{m}\htmlData{tutor-start=87,tutor-end=88}{n}} \htmlData{tutor-start=90,tutor-end=96}{\cdot }\frac{\htmlData{tutor-start=102,tutor-end=103}{m}^{\htmlData{tutor-start=105,tutor-end=106}{2}}\htmlData{tutor-start=107,tutor-end=108}{n}^{\htmlData{tutor-start=110,tutor-end=111}{2}}}{\htmlData{tutor-start=114,tutor-end=115}{(}\htmlData{tutor-start=115,tutor-end=116}{n}^{\htmlData{tutor-start=118,tutor-end=119}{2}}\htmlData{tutor-start=120,tutor-end=121}{-}\htmlData{tutor-start=121,tutor-end=122}{m}^{\htmlData{tutor-start=124,tutor-end=125}{2}}\htmlData{tutor-start=126,tutor-end=127}{)}^{\htmlData{tutor-start=129,tutor-end=130}{2}}}。约分：分子的 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} 与分母中的 m2n2\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{n}^{\htmlData{tutor-start=8,tutor-end=9}{2}} 约去一个 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n}，使得 m2n2\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{n}^{\htmlData{tutor-start=8,tutor-end=9}{2}} 变为 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} 并移至分子。最终结果为 mn(m2+n2)(n2−m2)2\frac{\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{m}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{n}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{n}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{m}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{)}^{\htmlData{tutor-start=38,tutor-end=39}{2}}}。同样需满足 m,n≠0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n} \neq \htmlData{tutor-start=9,tutor-end=10}{0} 且 n≠±m\htmlData{tutor-start=0,tutor-end=1}{n} \neq \htmlData{tutor-start=7,tutor-end=11}{\pm }\htmlData{tutor-start=11,tutor-end=12}{m}。

u+v(u−v)2=mn(m2+n2)(n2−m2)2\frac{\htmlData{tutor-start=6,tutor-end=7}{u}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}}{\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{u}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}\htmlData{tutor-start=21,tutor-end=22}{=}\frac{\htmlData{tutor-start=28,tutor-end=29}{m}\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{m}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{n}^{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{)}}{\htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{n}^{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{m}^{\htmlData{tutor-start=55,tutor-end=56}{2}}\htmlData{tutor-start=57,tutor-end=58}{)}^{\htmlData{tutor-start=60,tutor-end=61}{2}}}
相似题目从题目升维到题型

正在比较题型结构与解法路线...

4

例题解析 · 分式/因式分解/通分

计算：b−ca2−ab−ac+bc−c−ab2−bc−ab+ac+a−bc2−ac−bc+ab\frac{\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}}{\htmlData{tutor-start=11,tutor-end=12}{a}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{c}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{b}\htmlData{tutor-start=24,tutor-end=25}{c}}\htmlData{tutor-start=26,tutor-end=27}{-}\frac{\htmlData{tutor-start=33,tutor-end=34}{c}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{a}}{\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{b}\htmlData{tutor-start=45,tutor-end=46}{c}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{a}\htmlData{tutor-start=48,tutor-end=49}{b}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{a}\htmlData{tutor-start=51,tutor-end=52}{c}}\htmlData{tutor-start=53,tutor-end=54}{+}\frac{\htmlData{tutor-start=60,tutor-end=61}{a}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{b}}{\htmlData{tutor-start=65,tutor-end=66}{c}^{\htmlData{tutor-start=68,tutor-end=69}{2}}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{a}\htmlData{tutor-start=72,tutor-end=73}{c}\htmlData{tutor-start=73,tutor-end=74}{-}\htmlData{tutor-start=74,tutor-end=75}{b}\htmlData{tutor-start=75,tutor-end=76}{c}\htmlData{tutor-start=76,tutor-end=77}{+}\htmlData{tutor-start=77,tutor-end=78}{a}\htmlData{tutor-start=78,tutor-end=79}{b}}。

答案：-2/(a-c)

题目标签：分解分母后通分化简

知识点
解题操作与技能

解题过程

分解三个分母后通分

把循环分式和化成一个简单分式

(1)
分别分解分母

三个二次式都可以按分组法分成两个一次因式。

详细展开：观察原式中的三个分母，它们都是关于 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c} 的二次齐次多项式。为了进行通分运算，我们需要先将每个分母分解为一次因式的乘积。对于第一个分母 a2−ab−ac+bc\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{c}，我们采用分组分解法：前两项提取公因式 a\htmlData{tutor-start=0,tutor-end=1}{a} 得到 a(a−b)\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)}，后两项提取公因式 −c\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{c} 得到 −c(a−b)\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)}，合并后即为 (a−b)(a−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}。同理，第二个分母 b2−bc−ab+ac\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{c} 可以重组为 b(b−a)−c(b−a)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{)}，即 (b−a)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}；第三个分母 c2−ac−bc+ab\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{b} 可以重组为 c(c−a)−b(c−a)\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{)}，即 (c−a)(c−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}。这一步将复杂的二次分母转化为了简单的线性因子乘积形式。

a2−ab−ac+bc=(a−b)(a−c),b2−bc−ab+ac=(b−a)(b−c),c2−ac−bc+ab=(c−a)(c−b)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{a}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{,}\quad \htmlData{tutor-start=32,tutor-end=33}{b}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{b}\htmlData{tutor-start=39,tutor-end=40}{c}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{a}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{a}\htmlData{tutor-start=45,tutor-end=46}{c}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{b}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{a}\htmlData{tutor-start=51,tutor-end=52}{)}\htmlData{tutor-start=52,tutor-end=53}{(}\htmlData{tutor-start=53,tutor-end=54}{b}\htmlData{tutor-start=54,tutor-end=55}{-}\htmlData{tutor-start=55,tutor-end=56}{c}\htmlData{tutor-start=56,tutor-end=57}{)}\htmlData{tutor-start=57,tutor-end=58}{,}\quad \htmlData{tutor-start=64,tutor-end=65}{c}^{\htmlData{tutor-start=67,tutor-end=68}{2}}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{a}\htmlData{tutor-start=71,tutor-end=72}{c}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{b}\htmlData{tutor-start=74,tutor-end=75}{c}\htmlData{tutor-start=75,tutor-end=76}{+}\htmlData{tutor-start=76,tutor-end=77}{a}\htmlData{tutor-start=77,tutor-end=78}{b}\htmlData{tutor-start=78,tutor-end=79}{=}\htmlData{tutor-start=79,tutor-end=80}{(}\htmlData{tutor-start=80,tutor-end=81}{c}\htmlData{tutor-start=81,tutor-end=82}{-}\htmlData{tutor-start=82,tutor-end=83}{a}\htmlData{tutor-start=83,tutor-end=84}{)}\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{c}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{b}\htmlData{tutor-start=88,tutor-end=89}{)}
(2)
统一符号并通分

统一使用公共分母 (a−b)(a−c)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)}，注意第二、第三个分母中的负号。

详细展开：为了将三个分式合并，我们需要统一分母。观察上一步得到的因式：(a−b)(a−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}、(b−a)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)} 和 (c−a)(c−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}。利用相反数关系 b−a=−(a−b)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}，c−a=−(a−c)\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}，c−b=−(b−c)\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}，我们将所有分母统一转化为包含 (a−b),(b−c),(a−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{c}\htmlData{tutor-start=18,tutor-end=19}{)} 的形式。第一个分母不变；第二个分母 (b−a)(b−c)=−(a−b)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{b}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{c}\htmlData{tutor-start=23,tutor-end=24}{)}，前面的负号移到分式整体前，与原式中的减号结合变为加号，或者保留负号在分母中处理；第三个分母 (c−a)(c−b)=[−(a−c)][−(b−c)]=(a−c)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{[}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{c}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{]}\htmlData{tutor-start=21,tutor-end=22}{[}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{c}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{]} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{a}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{c}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{b}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{c}\htmlData{tutor-start=41,tutor-end=42}{)}。为了简便，我们选取最小公分母为 (a−b)(b−c)(a−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)}。第一项分子需乘 (b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{)}，第二项分母为 −(a−b)(b−c)\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{)}，通分时需注意符号变化，第三项分母为 (a−c)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}，分子需乘 (a−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}。经过仔细调整符号，原式转化为同分母分式相加减。

(b−c)2−(a−c)2+(a−b)2(a−b)(a−c)(b−c)\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{b}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}}}{\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{a}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{b}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{a}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{c}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{b}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{c}\htmlData{tutor-start=51,tutor-end=52}{)}}
(3)
化简分子

展开或继续使用平方差，分子为 −2(a−b)(b−c)\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{)}，与公共分母约分。

详细展开：现在我们需要化简分子 N=(b−c)2−(a−c)2+(a−b)2\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{)}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}} 并与分母约分。首先展开分子中的完全平方式：(b−c)2=b2−2bc+c2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}^{\htmlData{tutor-start=25,tutor-end=26}{2}}，(a−c)2=a2−2ac+c2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{a}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}^{\htmlData{tutor-start=25,tutor-end=26}{2}}，(a−b)2=a2−2ab+b2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{a}\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{b}^{\htmlData{tutor-start=25,tutor-end=26}{2}}。代入得 N=(b2−2bc+c2)−(a2−2ac+c2)+(a2−2ab+b2)\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{b}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{c}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{)} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{a}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{a}\htmlData{tutor-start=33,tutor-end=34}{c}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{c}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=43}{+} \htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{a}^{\htmlData{tutor-start=48,tutor-end=49}{2}}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{2}\htmlData{tutor-start=52,tutor-end=53}{a}\htmlData{tutor-start=53,tutor-end=54}{b}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{b}^{\htmlData{tutor-start=58,tutor-end=59}{2}}\htmlData{tutor-start=60,tutor-end=61}{)}。去括号并合并同类项：b2−2bc+c2−a2+2ac−c2+a2−2ab+b2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{c}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{c}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{c}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{a}\htmlData{tutor-start=40,tutor-end=41}{b}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{b}^{\htmlData{tutor-start=45,tutor-end=46}{2}}。其中 −a2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 与 +a2\htmlData{tutor-start=0,tutor-end=1}{+}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 抵消，+c2\htmlData{tutor-start=0,tutor-end=1}{+}\htmlData{tutor-start=1,tutor-end=2}{c}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 与 −c2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{c}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 抵消。剩余项为 2b2−2bc+2ac−2ab\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{c} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{b}。提取公因式 2\htmlData{tutor-start=0,tutor-end=1}{2} 得 2(b2−bc+ac−ab)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}^{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{c} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{b}\htmlData{tutor-start=22,tutor-end=23}{)}。对括号内进行分组分解：b(b−c)−a(b−c)=(b−a)(b−c)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{a}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{c}\htmlData{tutor-start=27,tutor-end=28}{)}。注意到 b−a=−(a−b)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{)}，所以分子 N=−2(a−b)(b−c)\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{)}。最后，将分子与分母 (a−b)(a−c)(b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)} 进行约分，消去公共因子 (a−b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)} 和 (b−c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{)}，得到最终结果 −2a−c\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{c}}，即 −2a−c\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{c}}。

−2(a−b)(b−c)(a−b)(a−c)(b−c)=−2a−c\frac{-2(a-b)(b-c)}{(a-b)(a-c)(b-c)}=-\frac2{a-c}
相似题目从题目升维到题型

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5

例题解析 · 分式/整体代入

若 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}，求 aab+a+1+bbc+b+1+cac+c+1\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{+}\frac{\htmlData{tutor-start=23,tutor-end=24}{b}}{\htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{c}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{b}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{1}}\htmlData{tutor-start=33,tutor-end=34}{+}\frac{\htmlData{tutor-start=40,tutor-end=41}{c}}{\htmlData{tutor-start=43,tutor-end=44}{a}\htmlData{tutor-start=44,tutor-end=45}{c}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{c}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{1}}。

答案：1\htmlData{tutor-start=0,tutor-end=1}{1}

题目标签：利用整体条件局部代入

知识点
解题操作与技能

解题过程

利用 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 局部改写分母

把三个分式拼成同分母并求和

(1)
改写第一个分式

用 1=abc\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{c} 替换第一个分母中的常数 1，再提出 a\htmlData{tutor-start=0,tutor-end=1}{a}。

详细展开：首先观察第一个分式 aab+a+1\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}} 的分母为 ab+a+1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}。题目给出的关键条件是 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}。为了将三个分式的分母统一，我们尝试利用这个条件对第一个分式的分母进行变形。我们将分母中的常数项 1\htmlData{tutor-start=0,tutor-end=1}{1} 替换为 abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}，此时分母变为 ab+a+abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c}。接着，我们观察新分母 ab+a+abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c} 的每一项都含有公因式 a\htmlData{tutor-start=0,tutor-end=1}{a}，因此可以提取公因式 a\htmlData{tutor-start=0,tutor-end=1}{a}，得到 a(b+1+bc)\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{)}。此时，原分式变为 aa(b+1+bc)\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{c}\htmlData{tutor-start=17,tutor-end=18}{)}}。由于 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 隐含 a≠0\htmlData{tutor-start=0,tutor-end=1}{a} \neq \htmlData{tutor-start=7,tutor-end=8}{0}，我们可以安全地约去分子和分母中的 a\htmlData{tutor-start=0,tutor-end=1}{a}，最终得到 1b+1+bc\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{c}}，即 1bc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}。这一步成功将第一个分式的分母转化为了目标形式。

aab+a+1=aab+a+abc=1bc+b+1\frac{a}{ab+a+1}=\frac{a}{ab+a+abc}=\frac1{bc+b+1}
(2)
改写第三个分式

第三项分子分母同乘 b\htmlData{tutor-start=0,tutor-end=1}{b}，再使用 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}。

详细展开：接下来处理第三个分式 cac+c+1\frac{\htmlData{tutor-start=6,tutor-end=7}{c}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{c}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}。我们的目标是将其分母也转化为 bc+b+1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}。观察发现，如果给分子和分母同时乘以 b\htmlData{tutor-start=0,tutor-end=1}{b}，分母中将出现 abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c} 项，这可以利用条件 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 进行简化。具体操作如下：分子 c\htmlData{tutor-start=0,tutor-end=1}{c} 乘以 b\htmlData{tutor-start=0,tutor-end=1}{b} 变为 bc\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}；分母 (ac+c+1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)} 乘以 b\htmlData{tutor-start=0,tutor-end=1}{b} 展开为 abc+bc+b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{c}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{b}。根据已知条件 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}，我们将分母中的 abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c} 替换为 1\htmlData{tutor-start=0,tutor-end=1}{1}，于是分母变为 1+bc+b\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{b}，即 bc+b+1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}。因此，第三个分式变形为 bcbc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c}}{\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}}。此时，三个分式的分母已经完全统一。

cac+c+1=bcabc+bc+b=bcbc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{c}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{c}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\htmlData{tutor-start=23,tutor-end=24}{b}\htmlData{tutor-start=24,tutor-end=25}{c}}{\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{b}\htmlData{tutor-start=29,tutor-end=30}{c}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{c}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{b}}\htmlData{tutor-start=36,tutor-end=37}{=}\frac{\htmlData{tutor-start=43,tutor-end=44}{b}\htmlData{tutor-start=44,tutor-end=45}{c}}{\htmlData{tutor-start=47,tutor-end=48}{b}\htmlData{tutor-start=48,tutor-end=49}{c}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{b}\htmlData{tutor-start=51,tutor-end=52}{+}\htmlData{tutor-start=52,tutor-end=53}{1}}
(3)
三个同分母分式相加

第一个、第二个和改写后的第三个分式拥有同一分母。

详细展开：现在，我们将三个分式相加。第一个分式已化为 1bc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}，第二个分式原本就是 bbc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{b}}{\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}}，第三个分式已化为 bcbc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{c}}{\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}}。由于它们的分母完全相同，均为 bc+b+1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}，我们可以直接将分子相加，分母保持不变。分子之和为 1+b+bc\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{c}。因此，总和为 1+b+bcbc+b+1\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}}{\htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{c}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{b}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1}}。观察发现，分子 1+b+bc\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{c} 与分母 bc+b+1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1} 完全相同。只要分母不为零，该分式的值即为 1。由 abc=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 知 b,c\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{c} 不为零且同号等，分母 bc+b+1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1} 在实数范围内通常不为零（除非特定复数情况，但在中学数学背景下默认为非零实数），故结果为 1。

1bc+b+1+bbc+b+1+bcbc+b+1=1\frac1{bc+b+1}+\frac b{bc+b+1}+\frac{bc}{bc+b+1}=1
相似题目从题目升维到题型

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6

例题解析 · 分式/比例性质/分类讨论

已知 x+y−zz=x−y+zy=−x+y+zx\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{z}}{\htmlData{tutor-start=13,tutor-end=14}{z}}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{y}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{z}}{\htmlData{tutor-start=29,tutor-end=30}{y}}\htmlData{tutor-start=31,tutor-end=32}{=}\frac{\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{x}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{y}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{z}}{\htmlData{tutor-start=46,tutor-end=47}{x}}，且 xyz≠0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{z}\ne\htmlData{tutor-start=6,tutor-end=7}{0}，求 (x+y)(y+z)(z+x)xyz\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{y}\htmlData{tutor-start=25,tutor-end=26}{z}} 的值。

答案：8 或 -1

题目标签：引入比例参数分类求值

知识点
解题操作与技能

解题过程

引入公共比例参数并分类

求循环乘积比的所有可能值

(1)
设置公共比值

设三个分式的公共值为 k\htmlData{tutor-start=0,tutor-end=1}{k}，再令 t=k+1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}，可把三个等式改写成完全对称的形式。

详细展开：首先观察已知条件中的三个分式，它们的分子都是两个变量之和减去第三个变量，而分母恰好是被减去的那个变量。为了利用这种对称性，我们设这三个分式的公共值为 k\htmlData{tutor-start=0,tutor-end=1}{k}。根据定义，有 x+y−zz=k\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{z}}{\htmlData{tutor-start=13,tutor-end=14}{z}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{k}，x−y+zy=k\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{z}}{\htmlData{tutor-start=13,tutor-end=14}{y}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{k}，−x+y+zx=k\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{z}}{\htmlData{tutor-start=14,tutor-end=15}{x}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{k}。接下来对每个等式进行变形：第一个等式两边同乘 z\htmlData{tutor-start=0,tutor-end=1}{z} 得 x+y−z=kz\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{z}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{z}，移项得 x+y=(k+1)z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{z}；同理，第二个等式变形为 x+z=(k+1)y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{y}；第三个等式变形为 y+z=(k+1)x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{x}。为了书写简洁并突出对称结构，令 t=k+1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}。此时，原方程组转化为完全对称的形式：x+y=tz\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{z}，x+z=ty\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{y}，y+z=tx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{x}。这一步将复杂的连等式转化为了线性方程组，且系数统一为参数 t\htmlData{tutor-start=0,tutor-end=1}{t}。

x+y=tz,x+z=ty,y+z=tx\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{z}\htmlData{tutor-start=6,tutor-end=7}{,}\quad \htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{z}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{t}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{,}\quad \htmlData{tutor-start=26,tutor-end=27}{y}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{z}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{t}\htmlData{tutor-start=31,tutor-end=32}{x}
(2)
确定参数的两种可能

前两式相减得到 (t+1)(y−z)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{y}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{z}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}，循环相减同理。若 t≠−1\htmlData{tutor-start=0,tutor-end=1}{t}\ne\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}，则 x=y=z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{z}，代入得 t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}；另一种情形是 t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}。

详细展开：利用上一步得到的对称方程组 x+y=tz\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{z} (1)，x+z=ty\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{y} (2)，y+z=tx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{x} (3) 来确定参数 t\htmlData{tutor-start=0,tutor-end=1}{t} 的可能值。我们将方程 (1) 与方程 (2) 相减：(x+y)−(x+z)=tz−ty\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{z}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{t}\htmlData{tutor-start=15,tutor-end=16}{z} \htmlData{tutor-start=17,tutor-end=18}{-} \htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{y}。左边化简为 y−z\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{z}，右边提取公因式 t\htmlData{tutor-start=0,tutor-end=1}{t} 得到 t(z−y)\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)}。注意到 z−y=−(y−z)\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{z}\htmlData{tutor-start=11,tutor-end=12}{)}，所以方程变为 y−z=−t(y−z)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{z} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{t}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{z}\htmlData{tutor-start=12,tutor-end=13}{)}。移项整理得 (y−z)+t(y−z)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{z}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{z}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0}，即 (t+1)(y−z)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{y}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{z}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}。同理，将方程 (2) 与 (3) 相减可得 (t+1)(z−x)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{z}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}；将方程 (3) 与 (1) 相减可得 (t+1)(x−y)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}。这就产生了两种情形：情形一，若 t+1≠0\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1} \ne \htmlData{tutor-start=8,tutor-end=9}{0}（即 t≠−1\htmlData{tutor-start=0,tutor-end=1}{t} \ne \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}），则必须有 y−z=0\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0}，z−x=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0}，x−y=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0}，这意味着 x=y=z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{z}。将 x=y=z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{z} 代入方程 (1) 得 2x=tx\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{x}，因 x≠0\htmlData{tutor-start=0,tutor-end=1}{x}\ne\htmlData{tutor-start=4,tutor-end=5}{0}，解得 t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。情形二，若 t+1=0\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{0}，即 t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}，此时上述乘积为零恒成立，无需 x,y,z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{z} 相等。综上，t\htmlData{tutor-start=0,tutor-end=1}{t} 的可能取值为 2\htmlData{tutor-start=0,tutor-end=1}{2} 或 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}。

t=2或t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\quad\text{\htmlData{tutor-start=14,tutor-end=15}{或}}\quad \htmlData{tutor-start=22,tutor-end=23}{t}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}
(3)
计算目标乘积

三个两两和分别为 tz,tx,ty\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{t}\htmlData{tutor-start=7,tutor-end=8}{y}，所以目标式等于 t3\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{3}}。

详细展开：我们需要计算的目标式是 (x+y)(y+z)(z+x)xyz\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{y}\htmlData{tutor-start=25,tutor-end=26}{z}}。回顾第一步中建立的对称关系：x+y=tz\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{z}，y+z=tx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{x}，z+x=ty\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{y}。我们将这三个等式直接代入目标式的分子中。分子变为 (tz)⋅(tx)⋅(ty)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{z}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=11}{\cdot }\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=22}{\cdot }\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{t}\htmlData{tutor-start=24,tutor-end=25}{y}\htmlData{tutor-start=25,tutor-end=26}{)}。根据乘法交换律和结合律，这等于 t3⋅xyz\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=12}{\cdot }\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{z}。因此，原式可以化简为 t3xyzxyz\frac{\htmlData{tutor-start=6,tutor-end=7}{t}^{\htmlData{tutor-start=9,tutor-end=10}{3}} \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{z}}{\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{z}}。由于题目已知 xyz≠0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{z} \ne \htmlData{tutor-start=8,tutor-end=9}{0}，我们可以安全地约去分子和分母中的 xyz\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{z}，得到结果 t3\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{3}}。根据上一步的结论，t\htmlData{tutor-start=0,tutor-end=1}{t} 有两个可能值：2\htmlData{tutor-start=0,tutor-end=1}{2} 和 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}。当 t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 时，目标值为 23=8\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{8}；当 t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 时，目标值为 (−1)3=−1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}^{\htmlData{tutor-start=6,tutor-end=7}{3}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}。因此，该表达式的所有可能值为 8\htmlData{tutor-start=0,tutor-end=1}{8} 或 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}。

(x+y)(y+z)(z+x)xyz=t3=8或−1\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{y}\htmlData{tutor-start=25,tutor-end=26}{z}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{t}^{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{8}\quad\text{\htmlData{tutor-start=46,tutor-end=47}{或}}\quad\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{1}
相似题目从题目升维到题型

正在比较题型结构与解法路线...

7

课堂练习 · 分式/倒数关系/方程组

已知 1=xyx+y\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{y}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{y}}，z=yzy+z\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}，3=xzx+z\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}，则 x=\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}____。

答案：按原卷条件无解；若第二个左端应为 2，则 x=12/5

题目标签：倒数关系求未知数

知识点
解题操作与技能

解题过程

核对原卷条件并说明疑似修正

判断原题是否自洽，并给出数字 2 假设下的答案

(1)
按原卷变量 z\htmlData{tutor-start=0,tutor-end=1}{z} 推导

由 z=yzy+z\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}} 交叉相乘得 z(y+z)=yz\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{z}，因此 z2=0\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，即 z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}。

详细展开：首先观察题目中给出的第二个等式 z=yzy+z\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}。这是一个关于变量 z\htmlData{tutor-start=0,tutor-end=1}{z} 和 y\htmlData{tutor-start=0,tutor-end=1}{y} 的方程，且 z\htmlData{tutor-start=0,tutor-end=1}{z} 同时出现在等式的左边和右边的分子中。为了简化这个分式方程，我们采用交叉相乘的方法消除分母。根据分式的基本性质，若 A=BC\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{B}}{\htmlData{tutor-start=13,tutor-end=14}{C}} 且 C≠0\htmlData{tutor-start=0,tutor-end=1}{C} \neq \htmlData{tutor-start=7,tutor-end=8}{0}，则 AC=B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{B}。在此处，我们将等式两边同时乘以 (y+z)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{z}\htmlData{tutor-start=4,tutor-end=5}{)}，得到 z(y+z)=yz\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{z}。接下来展开左边的括号，得到 zy+z2=yz\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{y} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{z}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{z}。观察等式两边，发现都含有项 yz\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{z}（即 zy\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{y}）。我们将等式两边的 yz\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{z} 同时减去，即移项处理：z2=yz−yz\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{z}，从而得到 z2=0\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{0}。根据实数的性质，只有当底数为 0 时，其平方才为 0，因此解得 z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}。这一步推导完全基于代数恒等变形，未引入额外假设。

z(y+z)=yz⇒z2=0⇒z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{z}\htmlData{tutor-start=9,tutor-end=21}{\Rightarrow }\htmlData{tutor-start=21,tutor-end=22}{z}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=40}{\Rightarrow }\htmlData{tutor-start=40,tutor-end=41}{z}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{0}
(2)
发现与第三个等式矛盾

代入 3=xzx+z\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}} 后右边为 0，与 3 矛盾，所以原卷条件无解。

详细展开：上一步我们已经推导出强制结论 z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}。现在需要检查这个结论是否与题目中的其他条件相容。我们将 z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 代入第三个已知等式 3=xzx+z\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}} 中进行验证。首先计算等式右边的值：分子为 x⋅z=x⋅0=0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{z} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\htmlData{tutor-start=20,tutor-end=21}{0} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{0}；分母为 x+z=x+0=x\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{z} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{0} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{x}。因此，右边的分式变为 0x\frac{\htmlData{tutor-start=6,tutor-end=7}{0}}{\htmlData{tutor-start=9,tutor-end=10}{x}}。只要 x≠0\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{0}，该分式的值即为 0。然而，等式的左边明确给出数值 3。这就导致了 3=0\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0} 的荒谬结论。即使考虑 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 的情况，分母为零使得分式无意义，依然无法等于 3。因此，由第二个条件推出的 z=0\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 与第三个条件直接矛盾。这意味着原题给出的三个条件不可能同时成立，即原题在数学上是自相矛盾的，无解。

3=x⋅0x+0=0矛盾\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=14}{\cdot}\htmlData{tutor-start=14,tutor-end=15}{0}}{\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{0}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{0}\quad\text{\htmlData{tutor-start=34,tutor-end=35}{矛}\htmlData{tutor-start=35,tutor-end=36}{盾}}
(3)
计算疑似数字 2 时的结果

若第二个左端本意为数字 2，取倒数得到关于 1x,1y,1z\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{,}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{y}}\htmlData{tutor-start=23,tutor-end=24}{,}\frac{\htmlData{tutor-start=30,tutor-end=31}{1}}{\htmlData{tutor-start=33,tutor-end=34}{z}} 的线性方程组。第一式加第三式再减第二式可求 1x\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{x}}。

详细展开：鉴于原题存在矛盾，题目提示我们考虑“数字 2 假设”，即推测原题第二个条件中的字母 z\htmlData{tutor-start=0,tutor-end=1}{z} 可能是印刷错误，本意应为数字 2。若将第二个条件修正为 2=yzy+z\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}，则三个条件变为：(1) 1=xyx+y\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{y}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{y}}，(2) 2=yzy+z\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{y}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}，(3) 3=xzx+z\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{z}}{\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{z}}。这类形如 k=aba+b\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{b}}{\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{b}} 的方程，可以通过取倒数转化为线性方程。对每个等式两边取倒数：由 (1) 得 11=x+yxy=1y+1x\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{y}}{\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{y}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{y}} \htmlData{tutor-start=43,tutor-end=44}{+} \frac{\htmlData{tutor-start=51,tutor-end=52}{1}}{\htmlData{tutor-start=54,tutor-end=55}{x}}；由 (2) 得 12=y+zyz=1z+1y\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{y}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{z}}{\htmlData{tutor-start=25,tutor-end=26}{y}\htmlData{tutor-start=26,tutor-end=27}{z}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{z}} \htmlData{tutor-start=43,tutor-end=44}{+} \frac{\htmlData{tutor-start=51,tutor-end=52}{1}}{\htmlData{tutor-start=54,tutor-end=55}{y}}；由 (3) 得 13=x+zxz=1z+1x\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{z}}{\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{z}} \htmlData{tutor-start=29,tutor-end=30}{=} \frac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{z}} \htmlData{tutor-start=43,tutor-end=44}{+} \frac{\htmlData{tutor-start=51,tutor-end=52}{1}}{\htmlData{tutor-start=54,tutor-end=55}{x}}。设 A=1x,B=1y,C=1z\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{B}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{y}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{C}\htmlData{tutor-start=31,tutor-end=32}{=}\frac{\htmlData{tutor-start=38,tutor-end=39}{1}}{\htmlData{tutor-start=41,tutor-end=42}{z}}，则得到线性方程组：A+B=1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}，B+C=12\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{=}\frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}，A+C=13\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{=}\frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{3}}。为了求 x\htmlData{tutor-start=0,tutor-end=1}{x}（即求 A\htmlData{tutor-start=0,tutor-end=1}{A}），我们将第一式和第三式相加：(A+B)+(A+C)=1+13=43\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1} \htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{3}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{4}}{\htmlData{tutor-start=43,tutor-end=44}{3}}，即 2A+B+C=43\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{B} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} \frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{3}}。再减去第二式 (B+C=12)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}，得 2A=43−12=8−36=56\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{4}}{\htmlData{tutor-start=14,tutor-end=15}{3}} \htmlData{tutor-start=17,tutor-end=18}{-} \frac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{=} \frac{\htmlData{tutor-start=39,tutor-end=40}{8}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{3}}{\htmlData{tutor-start=44,tutor-end=45}{6}} \htmlData{tutor-start=47,tutor-end=48}{=} \frac{\htmlData{tutor-start=55,tutor-end=56}{5}}{\htmlData{tutor-start=58,tutor-end=59}{6}}。解得 A=512\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{5}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{2}}。因为 A=1x\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}}，所以 x=1A=125\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{A}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{2}}{\htmlData{tutor-start=28,tutor-end=29}{5}}。

1x+1y=1,1y+1z=12,1x+1z=13⇒2x=56⇒x=125\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{y}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\quad \frac{\htmlData{tutor-start=38,tutor-end=39}{1}}{\htmlData{tutor-start=41,tutor-end=42}{y}}\htmlData{tutor-start=43,tutor-end=44}{+}\frac{\htmlData{tutor-start=50,tutor-end=51}{1}}{\htmlData{tutor-start=53,tutor-end=54}{z}}\htmlData{tutor-start=55,tutor-end=56}{=}\frac{\htmlData{tutor-start=62,tutor-end=63}{1}}{\htmlData{tutor-start=65,tutor-end=66}{2}}\htmlData{tutor-start=67,tutor-end=68}{,}\quad \frac{\htmlData{tutor-start=80,tutor-end=81}{1}}{\htmlData{tutor-start=83,tutor-end=84}{x}}\htmlData{tutor-start=85,tutor-end=86}{+}\frac{\htmlData{tutor-start=92,tutor-end=93}{1}}{\htmlData{tutor-start=95,tutor-end=96}{z}}\htmlData{tutor-start=97,tutor-end=98}{=}\frac{\htmlData{tutor-start=104,tutor-end=105}{1}}{\htmlData{tutor-start=107,tutor-end=108}{3}}\htmlData{tutor-start=109,tutor-end=121}{\Rightarrow }\frac{\htmlData{tutor-start=127,tutor-end=128}{2}}{\htmlData{tutor-start=130,tutor-end=131}{x}}\htmlData{tutor-start=132,tutor-end=133}{=}\frac{\htmlData{tutor-start=139,tutor-end=140}{5}}{\htmlData{tutor-start=142,tutor-end=143}{6}}\htmlData{tutor-start=144,tutor-end=156}{\Rightarrow }\htmlData{tutor-start=156,tutor-end=157}{x}\htmlData{tutor-start=157,tutor-end=158}{=}\frac{\htmlData{tutor-start=164,tutor-end=165}{1}\htmlData{tutor-start=165,tutor-end=166}{2}}{\htmlData{tutor-start=168,tutor-end=169}{5}}
相似题目从题目升维到题型

正在比较题型结构与解法路线...

8

课堂练习 · 分式/根式/多项式求值

若 x=19−43\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{4}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}，则 x4−6x3−2x2+18x+23x2−8x+15=\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{4}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{6}\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{x}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{8}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{3}}{\htmlData{tutor-start=34,tutor-end=35}{x}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{8}\htmlData{tutor-start=41,tutor-end=42}{x}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{5}}\htmlData{tutor-start=46,tutor-end=47}{=}____。

答案：按原卷约为 61.526006；若根号内系数应为 8，则答案为 5

题目标签：二次根式条件下的分式求值

知识点
解题操作与技能

解题过程

核对根式条件并分别计算

给出原卷条件下的精确值与疑似修正值

(1)
对分子作多项式除法

记分子为 N\htmlData{tutor-start=0,tutor-end=1}{N}、分母为 D\htmlData{tutor-start=0,tutor-end=1}{D}。先把分式写成多项式部分与余式部分。

详细展开：记分子为 N(x)=x4−6x3−2x2+18x+23\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{x}^{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{8}\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{3}，分母为 D(x)=x2−8x+15\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{8}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{5}。为了简化计算，我们对 N(x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 除以 D(x)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 进行多项式长除法。首先，用 N(x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的最高次项 x4\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} 除以 D(x)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的最高次项 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，得到商的第一项 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。将 x2D(x)=x4−8x3+15x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{4}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{8}\htmlData{tutor-start=20,tutor-end=21}{x}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{5}\htmlData{tutor-start=28,tutor-end=29}{x}^{\htmlData{tutor-start=31,tutor-end=32}{2}} 从 N(x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 中减去，得到余式 2x3−17x2+18x+23\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{3}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{7}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{3}。接着，用 2x3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{3}} 除以 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 得到 2x\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}，计算 2xD(x)=2x3−16x2+30x\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{6}\htmlData{tutor-start=19,tutor-end=20}{x}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{x} 并相减，得到 −x2−12x+23\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{3}。最后，用 −x2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 除以 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 得到 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}，计算 −1D(x)=−x2+8x−15\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{8}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{5} 并相减，得到最终余式 −20x+38\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{8}。因此，N(x)=(x2+2x−1)D(x)+(38−20x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{D}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{)} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{8}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{x}\htmlData{tutor-start=33,tutor-end=34}{)}。

N=(x2+2x−1)D+(38−20x),D=x2−8x+15\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{D}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{8}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{,}\quad \htmlData{tutor-start=31,tutor-end=32}{D}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{x}^{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{8}\htmlData{tutor-start=40,tutor-end=41}{x}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{5}
(2)
代入原卷根式

原卷给出 x2=19−43\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{4}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{3}}。直接代入并有理化，可得到下面的精确代数数；其近似值不是常见的简洁填空答案。

详细展开：根据原题条件 x=19−43\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{4}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}，我们有 x2=19−43\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{4}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}。将此关系代入上一步得到的简化表达式中。首先计算分母 D(x)=x2−8x+15=(19−43)−8x+15=34−43−8x\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{8}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{5} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{9}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{4}\sqrt{\htmlData{tutor-start=32,tutor-end=33}{3}}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{8}\htmlData{tutor-start=37,tutor-end=38}{x}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{5} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{3}\htmlData{tutor-start=45,tutor-end=46}{4}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{4}\sqrt{\htmlData{tutor-start=54,tutor-end=55}{3}}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{8}\htmlData{tutor-start=58,tutor-end=59}{x}。分子部分的余式为 38−20x\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{x}。整式部分为 x2+2x−1=(19−43)+2x−1=18−43+2x\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{9}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{4}\sqrt{\htmlData{tutor-start=24,tutor-end=25}{3}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{8}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{4}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{3}}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{x}。因此原式值为 18−43+2x+38−20x34−43−8x\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{8}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{x}}{\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{4}\sqrt{\htmlData{tutor-start=42,tutor-end=43}{3}}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{8}\htmlData{tutor-start=46,tutor-end=47}{x}}。为了得到精确值，我们需要对分式部分进行有理化或通分合并。经过繁琐但直接的代数运算（通分、合并同类项、利用 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 替换高次项），最终得到精确表达式 −2073+19583+(−444+4363)19−4339\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{8}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{4}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{6}\sqrt{\htmlData{tutor-start=40,tutor-end=41}{3}}\htmlData{tutor-start=42,tutor-end=43}{)}\sqrt{\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{9}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{4}\sqrt{\htmlData{tutor-start=59,tutor-end=60}{3}}}}{\htmlData{tutor-start=64,tutor-end=65}{3}\htmlData{tutor-start=65,tutor-end=66}{9}}。其数值约为 61.526006\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{6}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{6}。

ND=−2073+19583+(−444+4363)19−4339≈61.526006\frac{\htmlData{tutor-start=6,tutor-end=7}{N}}{\htmlData{tutor-start=9,tutor-end=10}{D}}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{9}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{8}\sqrt{\htmlData{tutor-start=34,tutor-end=35}{3}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{4}\htmlData{tutor-start=41,tutor-end=42}{4}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{4}\htmlData{tutor-start=44,tutor-end=45}{3}\htmlData{tutor-start=45,tutor-end=46}{6}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{)}\sqrt{\htmlData{tutor-start=61,tutor-end=62}{1}\htmlData{tutor-start=62,tutor-end=63}{9}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{4}\sqrt{\htmlData{tutor-start=71,tutor-end=72}{3}}}}{\htmlData{tutor-start=76,tutor-end=77}{3}\htmlData{tutor-start=77,tutor-end=78}{9}}\htmlData{tutor-start=79,tutor-end=86}{\approx}\htmlData{tutor-start=86,tutor-end=87}{6}\htmlData{tutor-start=87,tutor-end=88}{1}\htmlData{tutor-start=88,tutor-end=89}{.}\htmlData{tutor-start=89,tutor-end=90}{5}\htmlData{tutor-start=90,tutor-end=91}{2}\htmlData{tutor-start=91,tutor-end=92}{6}\htmlData{tutor-start=92,tutor-end=93}{0}\htmlData{tutor-start=93,tutor-end=94}{0}\htmlData{tutor-start=94,tutor-end=95}{6}
(3)
说明疑似系数 8 时的结果

若根号内本意是 19−83\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{8}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}，则 x=4−3\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{-}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}} 且 x2−8x+13=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}。又有 N−5D=(x2−8x+13)(x2+2x−4)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{x}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{)}，因此分式等于 5。

详细展开：考虑到填空题通常期望简洁整数答案，我们推测原题根号内系数可能有误。若假设 x=19−83\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{8}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}，注意到 19−83=16−83+3=(4−3)2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{8}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{6} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{8}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{3} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{4}\htmlData{tutor-start=38,tutor-end=39}{-}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{3}}\htmlData{tutor-start=47,tutor-end=48}{)}^{\htmlData{tutor-start=50,tutor-end=51}{2}}，故 x=4−3\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{-}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}（取正值）。此时 x\htmlData{tutor-start=0,tutor-end=1}{x} 满足方程 x−4=−3⇒x2−8x+16=3⇒x2−8x+13=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=26}{\Rightarrow }\htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{8}\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{6}\htmlData{tutor-start=37,tutor-end=38}{=}\htmlData{tutor-start=38,tutor-end=39}{3} \htmlData{tutor-start=40,tutor-end=52}{\Rightarrow }\htmlData{tutor-start=52,tutor-end=53}{x}^{\htmlData{tutor-start=55,tutor-end=56}{2}}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{8}\htmlData{tutor-start=59,tutor-end=60}{x}\htmlData{tutor-start=60,tutor-end=61}{+}\htmlData{tutor-start=61,tutor-end=62}{1}\htmlData{tutor-start=62,tutor-end=63}{3}\htmlData{tutor-start=63,tutor-end=64}{=}\htmlData{tutor-start=64,tutor-end=65}{0}。我们要计算的分式分母为 D=x2−8x+15=(x2−8x+13)+2=2\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{x}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{8}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{2}。对于分子，我们考察 N(x)−5D(x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}。计算发现 N(x)−5D(x)=(x2−8x+13)(x2+2x−4)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{x}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{8}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{x}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{4}\htmlData{tutor-start=37,tutor-end=38}{)}。因为 x2−8x+13=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}，所以 N(x)−5D(x)=0\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}，即 N(x)=5D(x)\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}。因此分式值为 5。

x=4−3⇒N−5D=0⇒ND=5\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{-}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=24}{\Rightarrow }\htmlData{tutor-start=24,tutor-end=25}{N}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{D}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=41}{\Rightarrow}\frac{\htmlData{tutor-start=47,tutor-end=48}{N}}{\htmlData{tutor-start=50,tutor-end=51}{D}}\htmlData{tutor-start=52,tutor-end=53}{=}\htmlData{tutor-start=53,tutor-end=54}{5}
相似题目从题目升维到题型

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9

课堂练习 · 分式/倒数关系/含参恒等变形

设 xx2−mx+1=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}，则 x3x6−m3x3+1=\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{3}}}{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{6}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{m}^{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}}\htmlData{tutor-start=32,tutor-end=33}{=}____。

答案：1/(3m^2-2)

题目标签：利用倒数关系降幂

知识点
解题操作与技能

解题过程

把已知分式转成倒数和关系

降幂并化简目标分式

(1)
由已知等式得到倒数和

交叉相乘得 x2−mx+1=x\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{x}。因题中分式有定义且等于 1，所以 x≠0\htmlData{tutor-start=0,tutor-end=1}{x}\ne\htmlData{tutor-start=4,tutor-end=5}{0}；两边除以 x\htmlData{tutor-start=0,tutor-end=1}{x}。

详细展开：首先观察已知等式 xx2−mx+1=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}。由于分式有意义且值为 1，分母不为零且分子不为零，故 x≠0\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{0}。我们将等式两边同时乘以分母 x2−mx+1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}，得到 x=x2−mx+1\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{m}\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1}。接下来进行移项整理，将右边的项全部移到左边或合并同类项：x2−mx−x+1=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{m}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{0}，即 x2−(m+1)x+1=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{m}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{0}。为了构造倒数和的形式，我们利用 x≠0\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{0} 这一条件，将方程两边同时除以 x\htmlData{tutor-start=0,tutor-end=1}{x}。此时方程变为 x2x−(m+1)xx+1x=0\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{m}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{x}}{\htmlData{tutor-start=32,tutor-end=33}{x}} \htmlData{tutor-start=35,tutor-end=36}{+} \frac{\htmlData{tutor-start=43,tutor-end=44}{1}}{\htmlData{tutor-start=46,tutor-end=47}{x}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{0}，化简后得到 x−(m+1)+1x=0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{m}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{+} \frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{x}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0}。最后将常数项移至等式右边，即可得到关键的不变量关系式：x+1x=m+1\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{m}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{1}。这一步将二次方程转化为了一次倒数和的形式，为后续计算高次幂奠定了基础。

x2−(m+1)x+1=0⇒x+1x=m+1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{m}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=28}{\Rightarrow }\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{+}\frac{\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{x}}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{m}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{1}
(2)
求三次幂与倒数三次幂之和

使用 (u+v)3=u3+v3+3uv(u+v)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{u}^{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{v}^{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{u}\htmlData{tutor-start=24,tutor-end=25}{v}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{u}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{v}\htmlData{tutor-start=29,tutor-end=30}{)}，这里 u=x,v=1/x\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{v}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{x} 且 uv=1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。

详细展开：我们需要计算目标式中出现的 x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} 相关项，因此需要求出 x3+1x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} 的值。利用完全立方公式 (a+b)3=a3+b3+3ab(a+b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{3}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{b}^{\htmlData{tutor-start=23,tutor-end=24}{3}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=31}{b}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{a}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{)}。令 a=x\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}，b=1x\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}}，则有 ab=x⋅1x=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=13}{\cdot }\frac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\htmlData{tutor-start=22,tutor-end=23}{x}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{1}。代入公式得：(x+1x)3=x3+1x3+3⋅1⋅(x+1x)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\frac{\htmlData{tutor-start=9,tutor-end=10}{1}}{\htmlData{tutor-start=12,tutor-end=13}{x}}\htmlData{tutor-start=14,tutor-end=15}{)}^{\htmlData{tutor-start=17,tutor-end=18}{3}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{x}^{\htmlData{tutor-start=25,tutor-end=26}{3}} \htmlData{tutor-start=28,tutor-end=29}{+} \frac{\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{x}^{\htmlData{tutor-start=42,tutor-end=43}{3}}} \htmlData{tutor-start=46,tutor-end=47}{+} \htmlData{tutor-start=48,tutor-end=49}{3} \htmlData{tutor-start=50,tutor-end=56}{\cdot }\htmlData{tutor-start=56,tutor-end=57}{1} \htmlData{tutor-start=58,tutor-end=64}{\cdot }\htmlData{tutor-start=64,tutor-end=65}{(}\htmlData{tutor-start=65,tutor-end=66}{x}\htmlData{tutor-start=66,tutor-end=67}{+}\frac{\htmlData{tutor-start=73,tutor-end=74}{1}}{\htmlData{tutor-start=76,tutor-end=77}{x}}\htmlData{tutor-start=78,tutor-end=79}{)}。我们将上一步得到的结论 x+1x=m+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{m}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1} 代入该恒等式。左边变为 (m+1)3\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{3}}，右边变为 x3+1x3+3(m+1)\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{3}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{m}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{)}。为了求出 x3+1x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}}，我们将 3(m+1)\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 移项到左边：x3+1x3=(m+1)3−3(m+1)\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{m}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{3}} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{3}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{m}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{)}。接下来展开并化简右边：(m+1)3=m3+3m2+3m+1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{m}^{\htmlData{tutor-start=15,tutor-end=16}{3}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{m}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=30}{3}\htmlData{tutor-start=30,tutor-end=31}{m} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{1}，而 3(m+1)=3m+3\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{m} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{3}。相减得：(m3+3m2+3m+1)−(3m+3)=m3+3m2+3m+1−3m−3=m3+3m2−2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}^{\htmlData{tutor-start=4,tutor-end=5}{3}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{m} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{3}\htmlData{tutor-start=30,tutor-end=31}{m} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{)} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{m}^{\htmlData{tutor-start=42,tutor-end=43}{3}} \htmlData{tutor-start=45,tutor-end=46}{+} \htmlData{tutor-start=47,tutor-end=48}{3}\htmlData{tutor-start=48,tutor-end=49}{m}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{+} \htmlData{tutor-start=56,tutor-end=57}{3}\htmlData{tutor-start=57,tutor-end=58}{m} \htmlData{tutor-start=59,tutor-end=60}{+} \htmlData{tutor-start=61,tutor-end=62}{1} \htmlData{tutor-start=63,tutor-end=64}{-} \htmlData{tutor-start=65,tutor-end=66}{3}\htmlData{tutor-start=66,tutor-end=67}{m} \htmlData{tutor-start=68,tutor-end=69}{-} \htmlData{tutor-start=70,tutor-end=71}{3} \htmlData{tutor-start=72,tutor-end=73}{=} \htmlData{tutor-start=74,tutor-end=75}{m}^{\htmlData{tutor-start=77,tutor-end=78}{3}} \htmlData{tutor-start=80,tutor-end=81}{+} \htmlData{tutor-start=82,tutor-end=83}{3}\htmlData{tutor-start=83,tutor-end=84}{m}^{\htmlData{tutor-start=86,tutor-end=87}{2}} \htmlData{tutor-start=89,tutor-end=90}{-} \htmlData{tutor-start=91,tutor-end=92}{2}。因此，x3+1x3=m3+3m2−2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{3}} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{m}^{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{-} \htmlData{tutor-start=43,tutor-end=44}{2}。

x3+1x3=(m+1)3−3(m+1)=m3+3m2−2x^{3}+\frac1{x^{3}}=(m+1)^{3}-3(m+1)=m^{3}+3m^{2}-2
(3)
目标式分子分母同除以 x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}}

目标分式化为三次幂与倒数三次幂的组合，再代入上一步结论。

详细展开：现在处理目标分式 x3x6−m3x3+1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{3}}}{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{6}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{m}^{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}}。为了利用上一步求得的 x3+1x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}}，我们需要对分式进行变形。由于 x≠0\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{0}，我们可以将分子和分母同时除以 x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}}。分子变为 x3x3=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{3}}}{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{3}}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{1}。分母变为 x6−m3x3+1x3=x6x3−m3x3x3+1x3=x3−m3+1x3\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{6}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{m}^{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{3}}} \htmlData{tutor-start=33,tutor-end=34}{=} \frac{\htmlData{tutor-start=41,tutor-end=42}{x}^{\htmlData{tutor-start=44,tutor-end=45}{6}}}{\htmlData{tutor-start=48,tutor-end=49}{x}^{\htmlData{tutor-start=51,tutor-end=52}{3}}} \htmlData{tutor-start=55,tutor-end=56}{-} \frac{\htmlData{tutor-start=63,tutor-end=64}{m}^{\htmlData{tutor-start=66,tutor-end=67}{3}}\htmlData{tutor-start=68,tutor-end=69}{x}^{\htmlData{tutor-start=71,tutor-end=72}{3}}}{\htmlData{tutor-start=75,tutor-end=76}{x}^{\htmlData{tutor-start=78,tutor-end=79}{3}}} \htmlData{tutor-start=82,tutor-end=83}{+} \frac{\htmlData{tutor-start=90,tutor-end=91}{1}}{\htmlData{tutor-start=93,tutor-end=94}{x}^{\htmlData{tutor-start=96,tutor-end=97}{3}}} \htmlData{tutor-start=100,tutor-end=101}{=} \htmlData{tutor-start=102,tutor-end=103}{x}^{\htmlData{tutor-start=105,tutor-end=106}{3}} \htmlData{tutor-start=108,tutor-end=109}{-} \htmlData{tutor-start=110,tutor-end=111}{m}^{\htmlData{tutor-start=113,tutor-end=114}{3}} \htmlData{tutor-start=116,tutor-end=117}{+} \frac{\htmlData{tutor-start=124,tutor-end=125}{1}}{\htmlData{tutor-start=127,tutor-end=128}{x}^{\htmlData{tutor-start=130,tutor-end=131}{3}}}。重新排列分母的项，得到 x3+1x3−m3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{3}}。此时，整个分式化为 1x3+1x3−m3\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{3}} \htmlData{tutor-start=15,tutor-end=16}{+} \frac{\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{3}}} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{m}^{\htmlData{tutor-start=38,tutor-end=39}{3}}}。我们将上一步计算出的结果 x3+1x3=m3+3m2−2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{3}}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{m}^{\htmlData{tutor-start=29,tutor-end=30}{3}} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{m}^{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{-} \htmlData{tutor-start=43,tutor-end=44}{2} 代入分母中。分母变为 (m3+3m2−2)−m3\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m}^{\htmlData{tutor-start=4,tutor-end=5}{3}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{m}^{\htmlData{tutor-start=26,tutor-end=27}{3}}。合并同类项，m3\htmlData{tutor-start=0,tutor-end=1}{m}^{\htmlData{tutor-start=3,tutor-end=4}{3}} 与 −m3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{m}^{\htmlData{tutor-start=4,tutor-end=5}{3}} 抵消，剩下 3m2−2\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{m}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{2}。因此，原分式的值为 13m2−2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{2}}。这就是最终的化简结果。

x3x6−m3x3+1=1x3+1x3−m3=13m2−2\frac{x^{3}}{x^{6}-m^{3}x^{3}+1}=\frac1{x^{3}+\frac1{x^{3}}-m^{3}}=\frac1{3m^{2}-2}
相似题目从题目升维到题型

正在比较题型结构与解法路线...

10

课堂练习 · 比例性质/分类讨论

若 abc≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\ne\htmlData{tutor-start=6,tutor-end=7}{0}，且 a+bc=b+ca=c+ab\frac{\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}}{\htmlData{tutor-start=11,tutor-end=12}{c}}\htmlData{tutor-start=13,tutor-end=14}{=}\frac{\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}}{\htmlData{tutor-start=25,tutor-end=26}{a}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{c}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{a}}{\htmlData{tutor-start=39,tutor-end=40}{b}}，则 (a+b)(b+c)(a+c)abc=\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{c}}\htmlData{tutor-start=27,tutor-end=28}{=}____。

答案：8 或 -1

题目标签：比例参数的两种情形

知识点
解题操作与技能

解题过程

引入比例参数并分类

把连等式参数化，再分类求目标乘积

(1)
直接设置公共比值

把三个分式的公共值直接记为 t\htmlData{tutor-start=0,tutor-end=1}{t}，去分母后得到循环对称方程组。

详细展开：连等式的作用就是说明三个分式等于同一个数，因此无需先设 k\htmlData{tutor-start=0,tutor-end=1}{k} 再换元。由 abc≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\ne\htmlData{tutor-start=6,tutor-end=7}{0}，三个分母都不为零，可以合法地分别乘去分母，得到 a+b=tc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{c}、b+c=ta\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{a}、c+a=tb\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{b}。这样既消除了分母，又保留了轮换对称结构。

a+bc=b+ca=c+ab=t⟹{a+b=tc,b+c=ta,c+a=tb.\frac{\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}}{\htmlData{tutor-start=11,tutor-end=12}{c}}\htmlData{tutor-start=13,tutor-end=14}{=}\frac{\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}}{\htmlData{tutor-start=25,tutor-end=26}{a}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{c}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{a}}{\htmlData{tutor-start=39,tutor-end=40}{b}}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{t}\htmlData{tutor-start=43,tutor-end=58}{\Longrightarrow}\begin{cases}\htmlData{tutor-start=71,tutor-end=72}{a}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{b}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{t}\htmlData{tutor-start=76,tutor-end=77}{c}\htmlData{tutor-start=77,tutor-end=78}{,}\\\htmlData{tutor-start=80,tutor-end=81}{b}\htmlData{tutor-start=81,tutor-end=82}{+}\htmlData{tutor-start=82,tutor-end=83}{c}\htmlData{tutor-start=83,tutor-end=84}{=}\htmlData{tutor-start=84,tutor-end=85}{t}\htmlData{tutor-start=85,tutor-end=86}{a}\htmlData{tutor-start=86,tutor-end=87}{,}\\\htmlData{tutor-start=89,tutor-end=90}{c}\htmlData{tutor-start=90,tutor-end=91}{+}\htmlData{tutor-start=91,tutor-end=92}{a}\htmlData{tutor-start=92,tutor-end=93}{=}\htmlData{tutor-start=93,tutor-end=94}{t}\htmlData{tutor-start=94,tutor-end=95}{b}\htmlData{tutor-start=95,tutor-end=96}{.}\end{cases}
(2)
分类确定 t\htmlData{tutor-start=0,tutor-end=1}{t}

循环相减得到 (t+1)(a−b)=(t+1)(b−c)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{b}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{0}，据此分类。

详细展开：相邻两式相减可以消去一个变量，并把差因式提出来。例如 (a+b)−(b+c)=tc−ta\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{t}\htmlData{tutor-start=16,tutor-end=17}{a}，即 (t+1)(a−c)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}。循环操作后，若 t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}，方程组存在非零解；若 t≠−1\htmlData{tutor-start=0,tutor-end=1}{t}\ne\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}，则 a=b=c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{c}，再代入 a+b=tc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{c}，利用 a≠0\htmlData{tutor-start=0,tutor-end=1}{a}\ne\htmlData{tutor-start=4,tutor-end=5}{0} 得 t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。

t=−1或t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\quad\text{\htmlData{tutor-start=15,tutor-end=16}{或}}\quad \htmlData{tutor-start=23,tutor-end=24}{t}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{2}
(3)
代入目标乘积

三个两两和正好分别是 tc,ta,tb\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{t}\htmlData{tutor-start=7,tutor-end=8}{b}，所以目标式可整体代换为 t3\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{3}}。

详细展开：不要分别求 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c}。直接将 (a+b)(b+c)(a+c)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)} 替换为 (tc)(ta)(tb)=t3abc\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{t}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{t}^{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{a}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{c}，再用 abc≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\ne\htmlData{tutor-start=6,tutor-end=7}{0} 约去分母。代入 t=−1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 或 t=2\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}，得到目标式的两个可能值 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}、8\htmlData{tutor-start=0,tutor-end=1}{8}。

(a+b)(b+c)(a+c)abc=t3=−1或8\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{c}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{t}^{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}\quad\text{\htmlData{tutor-start=47,tutor-end=48}{或}}\quad\htmlData{tutor-start=54,tutor-end=55}{8}
相似题目从题目升维到题型

正在比较题型结构与解法路线...

11

课堂练习 · 分式/局部代入

设 x,y,z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{z} 为有理数，且 x+y+z≠0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}\ne\htmlData{tutor-start=8,tutor-end=9}{0}，xy+z=a\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{z}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{a}，yz+x=b\frac{\htmlData{tutor-start=6,tutor-end=7}{y}}{\htmlData{tutor-start=9,tutor-end=10}{z}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{x}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{b}，zx+y=c\frac{\htmlData{tutor-start=6,tutor-end=7}{z}}{\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{y}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{c}，则 a1+a+b1+b+c1+c=\frac a{1+a}+\frac b{1+b}+\frac c{1+c}=____。

答案：1\htmlData{tutor-start=0,tutor-end=1}{1}

题目标签：局部代入化简循环分式

知识点
解题操作与技能

解题过程

把每项还原为 x,y,z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{z}

利用局部代入证明循环和为 1

(1)
化简第一项

由 a=xy+z\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}}{\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{z}}，把 1+a\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a} 通分。题设保证相关分母有定义。

详细展开：首先观察目标表达式中的第一项 a1+a\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{a}}。已知条件给出 a=xy+z\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}}{\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{z}}。为了将这一项转化为关于 x,y,z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{z} 的更简洁形式，我们需要对分母进行通分处理。根据分式加法法则，1+a=1+xy+z=y+zy+z+xy+z=x+y+zy+z\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=9}{+} \frac{\htmlData{tutor-start=16,tutor-end=17}{x}}{\htmlData{tutor-start=19,tutor-end=20}{y}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{z}} \htmlData{tutor-start=24,tutor-end=25}{=} \frac{\htmlData{tutor-start=32,tutor-end=33}{y}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{z}}{\htmlData{tutor-start=37,tutor-end=38}{y}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{z}} \htmlData{tutor-start=42,tutor-end=43}{+} \frac{\htmlData{tutor-start=50,tutor-end=51}{x}}{\htmlData{tutor-start=53,tutor-end=54}{y}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{z}} \htmlData{tutor-start=58,tutor-end=59}{=} \frac{\htmlData{tutor-start=66,tutor-end=67}{x}\htmlData{tutor-start=67,tutor-end=68}{+}\htmlData{tutor-start=68,tutor-end=69}{y}\htmlData{tutor-start=69,tutor-end=70}{+}\htmlData{tutor-start=70,tutor-end=71}{z}}{\htmlData{tutor-start=73,tutor-end=74}{y}\htmlData{tutor-start=74,tutor-end=75}{+}\htmlData{tutor-start=75,tutor-end=76}{z}}。这里利用了题设中隐含的条件 y+z≠0\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z} \neq \htmlData{tutor-start=9,tutor-end=10}{0}（因为 a\htmlData{tutor-start=0,tutor-end=1}{a} 有定义）。接下来，将分子 a\htmlData{tutor-start=0,tutor-end=1}{a} 和通分后的分母代入原式：a1+a=xy+zx+y+zy+z\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{a}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\frac{\htmlData{tutor-start=28,tutor-end=29}{x}}{\htmlData{tutor-start=31,tutor-end=32}{y}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{z}}}{\frac{\htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{y}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{z}}{\htmlData{tutor-start=50,tutor-end=51}{y}\htmlData{tutor-start=51,tutor-end=52}{+}\htmlData{tutor-start=52,tutor-end=53}{z}}}。根据繁分数的化简规则，分子分母同时乘以 y+z\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{z}，得到 xx+y+z\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{z}}。这一步成功将含参数 a\htmlData{tutor-start=0,tutor-end=1}{a} 的项转化为了含变量 x,y,z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{z} 且分母为总和的形式。

a1+a=xy+z1+xy+z=xx+y+z\frac a{1+a}=\frac{\frac{x}{y+z}}{1+\frac{x}{y+z}}=\frac{x}{x+y+z}
(2)
循环得到另外两项

同理，把 b,c\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{c} 分别还原为 y,z\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{z} 对总和的占比。

详细展开：基于第一步的推导经验，我们观察到 b\htmlData{tutor-start=0,tutor-end=1}{b} 和 c\htmlData{tutor-start=0,tutor-end=1}{c} 的定义式与 a\htmlData{tutor-start=0,tutor-end=1}{a} 具有完全相同的循环对称结构。具体来说，b=yz+x\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{y}}{\htmlData{tutor-start=11,tutor-end=12}{z}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{x}}，c=zx+y\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{z}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{y}}。因此，我们可以直接将第一步中的变量进行轮换替换，而无需重复繁琐的通分过程。对于第二项，将 x\htmlData{tutor-start=0,tutor-end=1}{x} 替换为 y\htmlData{tutor-start=0,tutor-end=1}{y}，y\htmlData{tutor-start=0,tutor-end=1}{y} 替换为 z\htmlData{tutor-start=0,tutor-end=1}{z}，z\htmlData{tutor-start=0,tutor-end=1}{z} 替换为 x\htmlData{tutor-start=0,tutor-end=1}{x}（注意分母中的对应关系），可得 b1+b=yy+z+x=yx+y+z\frac{\htmlData{tutor-start=6,tutor-end=7}{b}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{y}}{\htmlData{tutor-start=25,tutor-end=26}{y}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{z}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{x}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{y}}{\htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{y}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{z}}。同理，对于第三项，将 x\htmlData{tutor-start=0,tutor-end=1}{x} 替换为 z\htmlData{tutor-start=0,tutor-end=1}{z}，y\htmlData{tutor-start=0,tutor-end=1}{y} 替换为 x\htmlData{tutor-start=0,tutor-end=1}{x}，z\htmlData{tutor-start=0,tutor-end=1}{z} 替换为 y\htmlData{tutor-start=0,tutor-end=1}{y}，可得 c1+c=zz+x+y=zx+y+z\frac{\htmlData{tutor-start=6,tutor-end=7}{c}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{c}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{z}}{\htmlData{tutor-start=25,tutor-end=26}{z}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{y}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{z}}{\htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{y}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{z}}。这里再次确认了所有分母均为 x+y+z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}，且由题设知该值不为零，因此这两个分式均有意义。

b1+b=yx+y+z,c1+c=zx+y+z\frac b{1+b}=\frac{y}{x+y+z},\quad \frac c{1+c}=\frac{z}{x+y+z}
(3)
相加

三个分式分母相同，分子和正好等于分母。

详细展开：现在我们将前三步得到的三个分式相加。原式变为 xx+y+z+yx+y+z+zx+y+z\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{y}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{z}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{y}}{\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{y}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{z}} \htmlData{tutor-start=34,tutor-end=35}{+} \frac{\htmlData{tutor-start=42,tutor-end=43}{z}}{\htmlData{tutor-start=45,tutor-end=46}{x}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{y}\htmlData{tutor-start=48,tutor-end=49}{+}\htmlData{tutor-start=49,tutor-end=50}{z}}。由于这三个分式的分母完全相同，均为 x+y+z\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z}，我们可以根据同分母分式加法法则，保持分母不变，将分子相加。即：原式 =x+y+zx+y+z\htmlData{tutor-start=0,tutor-end=1}{=} \frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{z}}{\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{z}}。根据题设条件 x+y+z≠0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{z} \neq \htmlData{tutor-start=11,tutor-end=12}{0}，我们可以安全地进行约分。分子和分母完全相同，因此比值为 1。最终得出结论：a1+a+b1+b+c1+c=1\frac a{1+a}+\frac b{1+b}+\frac c{1+c} = 1。整个过程逻辑闭环，每一步变形均有理有据。

x+y+zx+y+z=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{z}}{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{y}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{z}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}
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12

课堂练习 · 分式/整体代入/恒等变形

已知 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c} 均不为 0\htmlData{tutor-start=0,tutor-end=1}{0}，且 a+b+c=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，则 1b2+c2−a2+1c2+a2−b2+1a2+b2−c2=\frac1{b^{2}+c^{2}-a^{2}}+\frac1{c^{2}+a^{2}-b^{2}}+\frac1{a^{2}+b^{2}-c^{2}}=____。

答案：0\htmlData{tutor-start=0,tutor-end=1}{0}

题目标签：整体条件化简平方差分母

知识点
解题操作与技能

解题过程

利用和为零改写三个分母

通分证明循环分式和为 0

(1)
化简第一个分母

由 a+b+c=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0} 得 a=−(b+c)\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{)}，代入第一个分母。

详细展开：首先观察第一个分母 b2+c2−a2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}}。题目给出的核心条件是 a+b+c=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，这意味着我们可以将任意一个变量表示为另外两个变量之和的相反数。为了简化含有 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 的项，我们将条件变形为 a=−(b+c)\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{)}。接下来，将这个表达式代入分母中的 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。根据完全平方公式 (x+y)2=x2+2xy+y2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{x}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{y}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{y}^{\htmlData{tutor-start=25,tutor-end=26}{2}}，我们有 a2=(−(b+c))2=(b+c)2=b2+2bc+c2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{c}\htmlData{tutor-start=27,tutor-end=28}{)}^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{b}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{2}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{c}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{c}^{\htmlData{tutor-start=48,tutor-end=49}{2}}。现在进行代换：原分母变为 b2+c2−(b2+2bc+c2)\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{b}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{c}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{)}。去括号时要注意符号变化，即 b2+c2−b2−2bc−c2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{c}^{\htmlData{tutor-start=25,tutor-end=26}{2}}。合并同类项后，b2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 与 −b2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{b}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 抵消，c2\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 与 −c2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{c}^{\htmlData{tutor-start=4,tutor-end=5}{2}} 抵消，最终只剩下 −2bc\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{c}。这一步成功地将二次多项式分母简化为单项式乘积。

b2+c2−a2=b2+c2−(b+c)2=−2bc\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{b}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{c}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{c}\htmlData{tutor-start=34,tutor-end=35}{)}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{2}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{c}
(2)
循环化简另外两个分母

同理分别用 b=−(c+a)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{)}、c=−(a+b)\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{)}。

详细展开：由于题目中的表达式具有轮换对称性（即变量 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c} 的地位是平等的），我们可以直接利用第一步的结论，通过轮换变量来得到另外两个分母的简化结果，而无需重复繁琐的计算过程。对于第二个分母 c2+a2−b2\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}，我们将条件变形为 b=−(c+a)\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{)}。类似于第一步的推导，b2=(c+a)2=c2+2ca+a2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{c}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{c}\htmlData{tutor-start=28,tutor-end=29}{a}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{a}^{\htmlData{tutor-start=33,tutor-end=34}{2}}。代入分母得 c2+a2−(c2+2ca+a2)=c2+a2−c2−2ca−a2=−2ca\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{c}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{a}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{a}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{)} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{c}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{a}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{c}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{c}\htmlData{tutor-start=52,tutor-end=53}{a}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{a}^{\htmlData{tutor-start=57,tutor-end=58}{2}} \htmlData{tutor-start=60,tutor-end=61}{=} \htmlData{tutor-start=62,tutor-end=63}{-}\htmlData{tutor-start=63,tutor-end=64}{2}\htmlData{tutor-start=64,tutor-end=65}{c}\htmlData{tutor-start=65,tutor-end=66}{a}。同理，对于第三个分母 a2+b2−c2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{c}^{\htmlData{tutor-start=15,tutor-end=16}{2}}，利用 c=−(a+b)\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{)}，可得 c2=(a+b)2=a2+2ab+b2\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{b}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{b}^{\htmlData{tutor-start=33,tutor-end=34}{2}}。代入分母得 a2+b2−(a2+2ab+b2)=a2+b2−a2−2ab−b2=−2ab\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{b}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{b}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{)} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{a}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{a}\htmlData{tutor-start=52,tutor-end=53}{b}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{b}^{\htmlData{tutor-start=57,tutor-end=58}{2}} \htmlData{tutor-start=60,tutor-end=61}{=} \htmlData{tutor-start=62,tutor-end=63}{-}\htmlData{tutor-start=63,tutor-end=64}{2}\htmlData{tutor-start=64,tutor-end=65}{a}\htmlData{tutor-start=65,tutor-end=66}{b}。至此，三个分母分别被简化为 −2bc\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{c}、−2ca\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{a} 和 −2ab\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{b}。

c2+a2−b2=−2ca,a2+b2−c2=−2ab\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{a}\htmlData{tutor-start=22,tutor-end=23}{,}\quad \htmlData{tutor-start=29,tutor-end=30}{a}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{b}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{c}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{a}\htmlData{tutor-start=50,tutor-end=51}{b}
(3)
通分并使用整体条件

因为 abc≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}\ne\htmlData{tutor-start=6,tutor-end=7}{0}，可以通分到 2abc\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{c}；分子就是题设给出的和。

详细展开：现在将简化后的分母代回原式。原式变为 1−2bc+1−2ca+1−2ab\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{c}} \htmlData{tutor-start=15,tutor-end=16}{+} \frac{\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{c}\htmlData{tutor-start=29,tutor-end=30}{a}} \htmlData{tutor-start=32,tutor-end=33}{+} \frac{\htmlData{tutor-start=40,tutor-end=41}{1}}{\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{b}}。提取公因数 −12\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}，得到 −12(1bc+1ca+1ab)\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \left( \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{b}\htmlData{tutor-start=30,tutor-end=31}{c}} \htmlData{tutor-start=33,tutor-end=34}{+} \frac{\htmlData{tutor-start=41,tutor-end=42}{1}}{\htmlData{tutor-start=44,tutor-end=45}{c}\htmlData{tutor-start=45,tutor-end=46}{a}} \htmlData{tutor-start=48,tutor-end=49}{+} \frac{\htmlData{tutor-start=56,tutor-end=57}{1}}{\htmlData{tutor-start=59,tutor-end=60}{a}\htmlData{tutor-start=60,tutor-end=61}{b}} \right)。接下来进行通分，这三个分式的最简公分母是 abc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c}。第一项分子分母同乘 a\htmlData{tutor-start=0,tutor-end=1}{a} 得 aabc\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}}，第二项同乘 b\htmlData{tutor-start=0,tutor-end=1}{b} 得 babc\frac{\htmlData{tutor-start=6,tutor-end=7}{b}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}}，第三项同乘 c\htmlData{tutor-start=0,tutor-end=1}{c} 得 cabc\frac{\htmlData{tutor-start=6,tutor-end=7}{c}}{\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{c}}。于是括号内变为 a+b+cabc\frac{\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{c}}{\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{c}}。结合前面的系数，整个表达式等于 −12⋅a+b+cabc=−a+b+c2abc\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=19}{\cdot }\frac{\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{b}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{c}}{\htmlData{tutor-start=32,tutor-end=33}{a}\htmlData{tutor-start=33,tutor-end=34}{b}\htmlData{tutor-start=34,tutor-end=35}{c}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{-}\frac{\htmlData{tutor-start=46,tutor-end=47}{a}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{b}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{c}}{\htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=55}{a}\htmlData{tutor-start=55,tutor-end=56}{b}\htmlData{tutor-start=56,tutor-end=57}{c}}。最后，利用题目最初给出的条件 a+b+c=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，分子为 0。因为 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c} 均不为 0，所以分母 2abc≠0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{c} \neq \htmlData{tutor-start=10,tutor-end=11}{0}，故整个分式的值为 0。

−12bc−12ca−12ab=−a+b+c2abc=0-\frac1{2bc}-\frac1{2ca}-\frac1{2ab}=-\frac{a+b+c}{2abc}=0
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