2008 年高考数学（新课标卷理科）

本步分别计算表达式中的 $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$、$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{z}$ 以及分母 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}$ 的具体复数值，为后续除法做准备。

详细展开： 1. 已知 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \text{\htmlData{tutor-start=14,tutor-end=15}{i}}$。 2. 计算 $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}}$： $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{-} \text{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{1}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{-} \htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=44}{\cdot }\htmlData{tutor-start=44,tutor-end=45}{1} \htmlData{tutor-start=46,tutor-end=52}{\cdot }\text{\htmlData{tutor-start=58,tutor-end=59}{i}} \htmlData{tutor-start=61,tutor-end=62}{+} \htmlData{tutor-start=63,tutor-end=64}{(}\text{\htmlData{tutor-start=70,tutor-end=71}{i}}\htmlData{tutor-start=72,tutor-end=73}{)}^{\htmlData{tutor-start=75,tutor-end=76}{2}} \htmlData{tutor-start=78,tutor-end=79}{=} \htmlData{tutor-start=80,tutor-end=81}{1} \htmlData{tutor-start=82,tutor-end=83}{-} \htmlData{tutor-start=84,tutor-end=85}{2}\text{\htmlData{tutor-start=91,tutor-end=92}{i}} \htmlData{tutor-start=94,tutor-end=95}{-} \htmlData{tutor-start=96,tutor-end=97}{1} \htmlData{tutor-start=98,tutor-end=99}{=} \htmlData{tutor-start=100,tutor-end=101}{-}\htmlData{tutor-start=101,tutor-end=102}{2}\text{\htmlData{tutor-start=108,tutor-end=109}{i}}$。 3. 计算 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{z}$： $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{z} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1} \htmlData{tutor-start=9,tutor-end=10}{-} \text{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2} \htmlData{tutor-start=25,tutor-end=26}{-} \htmlData{tutor-start=27,tutor-end=28}{2}\text{\htmlData{tutor-start=34,tutor-end=35}{i}}$。 4. 计算分子 $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{z}$： $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{z} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\text{\htmlData{tutor-start=22,tutor-end=23}{i}}\htmlData{tutor-start=24,tutor-end=25}{)} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{2} \htmlData{tutor-start=31,tutor-end=32}{-} \htmlData{tutor-start=33,tutor-end=34}{2}\text{\htmlData{tutor-start=40,tutor-end=41}{i}}\htmlData{tutor-start=42,tutor-end=43}{)} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{2}\text{\htmlData{tutor-start=54,tutor-end=55}{i}} \htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=60}{2} \htmlData{tutor-start=61,tutor-end=62}{+} \htmlData{tutor-start=63,tutor-end=64}{2}\text{\htmlData{tutor-start=70,tutor-end=71}{i}} \htmlData{tutor-start=73,tutor-end=74}{=} \htmlData{tutor-start=75,tutor-end=76}{-}\htmlData{tutor-start=76,tutor-end=77}{2}$。 5. 计算分母 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1}$： $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{-} \text{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{1} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{-}\text{\htmlData{tutor-start=36,tutor-end=37}{i}}$。

本步将简化后的分子和分母相除，利用复数除法规则得出最终实数或复数结果。

详细展开： 1. 原式化为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{-}\text{\htmlData{tutor-start=17,tutor-end=18}{i}}}$。 2. 负号相消，得 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\text{\htmlData{tutor-start=15,tutor-end=16}{i}}}$。 3. 为了消除分母中的虚数单位，分子分母同时乘以 $\htmlData{tutor-start=0,tutor-end=1}{-}\text{\htmlData{tutor-start=7,tutor-end=8}{i}}$（或利用 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\text{\htmlData{tutor-start=15,tutor-end=16}{i}}} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{-}\text{\htmlData{tutor-start=28,tutor-end=29}{i}}$）： $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\text{\htmlData{tutor-start=15,tutor-end=16}{i}}} \htmlData{tutor-start=19,tutor-end=20}{=} \frac{\htmlData{tutor-start=27,tutor-end=28}{2} \htmlData{tutor-start=29,tutor-end=35}{\cdot }\htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{-}\text{\htmlData{tutor-start=43,tutor-end=44}{i}}\htmlData{tutor-start=45,tutor-end=46}{)}}{\text{\htmlData{tutor-start=54,tutor-end=55}{i}} \htmlData{tutor-start=57,tutor-end=63}{\cdot }\htmlData{tutor-start=63,tutor-end=64}{(}\htmlData{tutor-start=64,tutor-end=65}{-}\text{\htmlData{tutor-start=71,tutor-end=72}{i}}\htmlData{tutor-start=73,tutor-end=74}{)}} \htmlData{tutor-start=76,tutor-end=77}{=} \frac{\htmlData{tutor-start=84,tutor-end=85}{-}\htmlData{tutor-start=85,tutor-end=86}{2}\text{\htmlData{tutor-start=92,tutor-end=93}{i}}}{\htmlData{tutor-start=96,tutor-end=97}{-}\text{\htmlData{tutor-start=103,tutor-end=104}{i}}^{\htmlData{tutor-start=107,tutor-end=108}{2}}} \htmlData{tutor-start=111,tutor-end=112}{=} \frac{\htmlData{tutor-start=119,tutor-end=120}{-}\htmlData{tutor-start=120,tutor-end=121}{2}\text{\htmlData{tutor-start=127,tutor-end=128}{i}}}{\htmlData{tutor-start=131,tutor-end=132}{-}\htmlData{tutor-start=132,tutor-end=133}{(}\htmlData{tutor-start=133,tutor-end=134}{-}\htmlData{tutor-start=134,tutor-end=135}{1}\htmlData{tutor-start=135,tutor-end=136}{)}} \htmlData{tutor-start=138,tutor-end=139}{=} \frac{\htmlData{tutor-start=146,tutor-end=147}{-}\htmlData{tutor-start=147,tutor-end=148}{2}\text{\htmlData{tutor-start=154,tutor-end=155}{i}}}{\htmlData{tutor-start=158,tutor-end=159}{1}} \htmlData{tutor-start=161,tutor-end=162}{=} \htmlData{tutor-start=163,tutor-end=164}{-}\htmlData{tutor-start=164,tutor-end=165}{2}\text{\htmlData{tutor-start=171,tutor-end=172}{i}}$。

*等等，让我重新检查计算*。

重新核算步骤1中的分子： $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{z} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{z}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{z}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}$。 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\text{\htmlData{tutor-start=14,tutor-end=15}{i}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{-}\text{\htmlData{tutor-start=30,tutor-end=31}{i}}$。 $\htmlData{tutor-start=0,tutor-end=1}{z}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\text{\htmlData{tutor-start=18,tutor-end=19}{i}}\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}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\text{\htmlData{tutor-start=31,tutor-end=32}{i}}\htmlData{tutor-start=33,tutor-end=34}{)} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{-}\text{\htmlData{tutor-start=47,tutor-end=48}{i}}\htmlData{tutor-start=49,tutor-end=50}{)}\htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{+}\text{\htmlData{tutor-start=59,tutor-end=60}{i}}\htmlData{tutor-start=61,tutor-end=62}{)} \htmlData{tutor-start=63,tutor-end=64}{=} \htmlData{tutor-start=65,tutor-end=66}{-}\htmlData{tutor-start=66,tutor-end=67}{(}\htmlData{tutor-start=67,tutor-end=68}{1}^{\htmlData{tutor-start=70,tutor-end=71}{2}} \htmlData{tutor-start=73,tutor-end=74}{-} \text{\htmlData{tutor-start=81,tutor-end=82}{i}}^{\htmlData{tutor-start=85,tutor-end=86}{2}}\htmlData{tutor-start=87,tutor-end=88}{)} \htmlData{tutor-start=89,tutor-end=90}{=} \htmlData{tutor-start=91,tutor-end=92}{-}\htmlData{tutor-start=92,tutor-end=93}{(}\htmlData{tutor-start=93,tutor-end=94}{1} \htmlData{tutor-start=95,tutor-end=96}{-} \htmlData{tutor-start=97,tutor-end=98}{(}\htmlData{tutor-start=98,tutor-end=99}{-}\htmlData{tutor-start=99,tutor-end=100}{1}\htmlData{tutor-start=100,tutor-end=101}{)}\htmlData{tutor-start=101,tutor-end=102}{)} \htmlData{tutor-start=103,tutor-end=104}{=} \htmlData{tutor-start=105,tutor-end=106}{-}\htmlData{tutor-start=106,tutor-end=107}{2}$。分子确实是 -2。

重新核算分母： $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{-}\text{\htmlData{tutor-start=13,tutor-end=14}{i}}$。

除法： $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{-}\text{\htmlData{tutor-start=17,tutor-end=18}{i}}} \htmlData{tutor-start=21,tutor-end=22}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{2}}{\text{\htmlData{tutor-start=38,tutor-end=39}{i}}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{2} \htmlData{tutor-start=46,tutor-end=52}{\cdot }\frac{\htmlData{tutor-start=58,tutor-end=59}{1}}{\text{\htmlData{tutor-start=67,tutor-end=68}{i}}} \htmlData{tutor-start=71,tutor-end=72}{=} \htmlData{tutor-start=73,tutor-end=74}{2}\htmlData{tutor-start=74,tutor-end=75}{(}\htmlData{tutor-start=75,tutor-end=76}{-}\text{\htmlData{tutor-start=82,tutor-end=83}{i}}\htmlData{tutor-start=84,tutor-end=85}{)} \htmlData{tutor-start=86,tutor-end=87}{=} \htmlData{tutor-start=88,tutor-end=89}{-}\htmlData{tutor-start=89,tutor-end=90}{2}\text{\htmlData{tutor-start=96,tutor-end=97}{i}}$。

查看选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{2}\text{\htmlData{tutor-start=7,tutor-end=8}{i}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$ (C) 2 (D) -2

我的计算结果是 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$，对应选项 (B)。

*再次审题与常见陷阱检查*： 有没有可能我看错了题目表达式？ 题目：$\frac{\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{z}}{\htmlData{tutor-start=16,tutor-end=17}{z}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}}$。 另一种思路： $\frac{\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{z}}{\htmlData{tutor-start=16,tutor-end=17}{z}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}} \htmlData{tutor-start=21,tutor-end=22}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{z}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{z}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{)}}{\htmlData{tutor-start=37,tutor-end=38}{z}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1}}$。 代入 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\text{\htmlData{tutor-start=10,tutor-end=11}{i}}$。 分子：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\text{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{-}\text{\htmlData{tutor-start=21,tutor-end=22}{i}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{)} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\text{\htmlData{tutor-start=38,tutor-end=39}{i}}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{-}\text{\htmlData{tutor-start=51,tutor-end=52}{i}}\htmlData{tutor-start=53,tutor-end=54}{)} \htmlData{tutor-start=55,tutor-end=56}{=} \htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{1} \htmlData{tutor-start=60,tutor-end=61}{-}\text{\htmlData{tutor-start=67,tutor-end=68}{i}} \htmlData{tutor-start=70,tutor-end=71}{+}\text{\htmlData{tutor-start=77,tutor-end=78}{i}} \htmlData{tutor-start=80,tutor-end=81}{+}\text{\htmlData{tutor-start=87,tutor-end=88}{i}}^{\htmlData{tutor-start=91,tutor-end=92}{2}} \htmlData{tutor-start=94,tutor-end=95}{=} \htmlData{tutor-start=96,tutor-end=97}{-}\htmlData{tutor-start=97,tutor-end=98}{1} \htmlData{tutor-start=99,tutor-end=100}{-}\htmlData{tutor-start=100,tutor-end=101}{1} \htmlData{tutor-start=102,tutor-end=103}{=} \htmlData{tutor-start=104,tutor-end=105}{-}\htmlData{tutor-start=105,tutor-end=106}{2}$。 分母：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{-}\text{\htmlData{tutor-start=22,tutor-end=23}{i}}$。 结果：$\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{-}\text{\htmlData{tutor-start=17,tutor-end=18}{i}}} \htmlData{tutor-start=21,tutor-end=22}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{2}}{\text{\htmlData{tutor-start=38,tutor-end=39}{i}}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{2}\text{\htmlData{tutor-start=52,tutor-end=53}{i}}$。

难道题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{z}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{z}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}}$? 不，题目写的是 $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{z}$。 难道 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{+}\text{\htmlData{tutor-start=10,tutor-end=11}{i}}$? 题目写的是 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$。

让我们检查是否有一种代数变形能直接看出结果： $\frac{\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{z}}{\htmlData{tutor-start=16,tutor-end=17}{z}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}} \htmlData{tutor-start=21,tutor-end=22}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{z}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{z}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{z}}{\htmlData{tutor-start=40,tutor-end=41}{z}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{z} \htmlData{tutor-start=49,tutor-end=50}{-} \frac{\htmlData{tutor-start=57,tutor-end=58}{z}}{\htmlData{tutor-start=60,tutor-end=61}{z}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{1}}$。 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{-}\text{\htmlData{tutor-start=12,tutor-end=13}{i}}$。 $\frac{\htmlData{tutor-start=6,tutor-end=7}{z}}{\htmlData{tutor-start=9,tutor-end=10}{z}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{-}\text{\htmlData{tutor-start=30,tutor-end=31}{i}}}{\htmlData{tutor-start=34,tutor-end=35}{-}\text{\htmlData{tutor-start=41,tutor-end=42}{i}}} \htmlData{tutor-start=45,tutor-end=46}{=} \frac{\htmlData{tutor-start=53,tutor-end=54}{1}}{\htmlData{tutor-start=56,tutor-end=57}{-}\text{\htmlData{tutor-start=63,tutor-end=64}{i}}} \htmlData{tutor-start=67,tutor-end=68}{+} \htmlData{tutor-start=69,tutor-end=70}{1} \htmlData{tutor-start=71,tutor-end=72}{=} \text{\htmlData{tutor-start=79,tutor-end=80}{i}} \htmlData{tutor-start=82,tutor-end=83}{+} \htmlData{tutor-start=84,tutor-end=85}{1}$。 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{+}\text{\htmlData{tutor-start=13,tutor-end=14}{i}}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\text{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{)} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{+}\text{\htmlData{tutor-start=43,tutor-end=44}{i}}\htmlData{tutor-start=45,tutor-end=46}{)} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{2}\text{\htmlData{tutor-start=57,tutor-end=58}{i}}$。

结果稳定为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$。对应选项 (B)。

*自我修正*：很多在线题库或记忆可能有偏差，但根据严格的数学推导，答案应为 (B)。然而，如果是经典真题，有时会有变体。例如若题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{z}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{z}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}}$ 或其他。但基于用户提供的文本，必须忠实于文本。

*再次确认选项*： (A) $\htmlData{tutor-start=0,tutor-end=1}{2}\text{\htmlData{tutor-start=7,tutor-end=8}{i}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$ (C) 2 (D) -2

如果答案是 (D) -2，那么分母必须是实数或者分子分母虚部抵消。这里分母纯虚，分子实数，结果必为纯虚数。所以 (C) (D) 排除。剩下 (A) (B)。符号取决于 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\text{\htmlData{tutor-start=8,tutor-end=9}{i}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\text{\htmlData{tutor-start=20,tutor-end=21}{i}}$。所以是 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\text{\htmlData{tutor-start=8,tutor-end=9}{i}}$。

因此，正确选项是 (B)。

本步通过设未知数，将文字描述的几何关系转化为代数方程，确定三角形三边的比例关系。

详细展开： 1. 设等腰三角形的底边长为 $\htmlData{tutor-start=0,tutor-end=1}{a}$，腰长为 $\htmlData{tutor-start=0,tutor-end=1}{b}$。则三边分别为 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{a}$。 2. 题目条件：“周长是底边长的 5 倍”。 3. 周长 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{b} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{a}$。 4. 建立方程：$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{a} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{a}$。 5. 移项整理：$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{a} \htmlData{tutor-start=8,tutor-end=23}{\implies b = 2a}$。 6. 因此，三角形的三边长之比为 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{:} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{b} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{a} \htmlData{tutor-start=14,tutor-end=15}{:} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a} \htmlData{tutor-start=19,tutor-end=20}{:} \htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{a} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{1} \htmlData{tutor-start=28,tutor-end=29}{:} \htmlData{tutor-start=30,tutor-end=31}{2} \htmlData{tutor-start=32,tutor-end=33}{:} \htmlData{tutor-start=34,tutor-end=35}{2}$。 7. 为方便计算，设 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$，则腰长 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$。三边为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{2}$。

本步针对顶角（即两腰夹角），应用余弦定理公式进行计算。

详细展开： 1. 设顶角为 $\htmlData{tutor-start=0,tutor-end=6}{\theta}$。顶角所对的边是底边 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$，邻边是两腰 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{c}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}$。 2. 根据余弦定理：$\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{c}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{b}\htmlData{tutor-start=26,tutor-end=27}{c} \cos \htmlData{tutor-start=33,tutor-end=39}{\theta}$。 3. 代入数值：$\htmlData{tutor-start=0,tutor-end=1}{1}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{2}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{2} \htmlData{tutor-start=26,tutor-end=32}{\cdot }\htmlData{tutor-start=32,tutor-end=33}{2} \htmlData{tutor-start=34,tutor-end=40}{\cdot }\htmlData{tutor-start=40,tutor-end=41}{2} \htmlData{tutor-start=42,tutor-end=48}{\cdot }\cos \htmlData{tutor-start=53,tutor-end=59}{\theta}$。 4. 化简：$\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{4} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{8} \cos \htmlData{tutor-start=19,tutor-end=25}{\theta}$。 5. $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{8} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{8} \cos \htmlData{tutor-start=15,tutor-end=21}{\theta}$。 6. 移项：$\htmlData{tutor-start=0,tutor-end=1}{8} \cos \htmlData{tutor-start=7,tutor-end=14}{\theta }\htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{8} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{1} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{7}$。 7. 解得：$\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{7}}{\htmlData{tutor-start=23,tutor-end=24}{8}}$。 8. 对比选项，(D) 为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{7}}{\htmlData{tutor-start=9,tutor-end=10}{8}}$。

本步利用等比数列的基本公式，将 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{4}}$ 都用首项 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 和公比 $\htmlData{tutor-start=0,tutor-end=1}{q}$ 表示出来。

详细展开： 1. 已知公比 $\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$。 2. 通项公式：$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}$。 所以，$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=20}{\cdot }\htmlData{tutor-start=20,tutor-end=21}{2}^{\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{1}}$。 3. 前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和公式：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{n}}\htmlData{tutor-start=27,tutor-end=28}{)}}{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{q}}$ （当 $\htmlData{tutor-start=0,tutor-end=1}{q} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$）。 所以，$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{2}^{\htmlData{tutor-start=25,tutor-end=26}{4}}\htmlData{tutor-start=27,tutor-end=28}{)}}{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{=} \frac{\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{1}}\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{6}\htmlData{tutor-start=53,tutor-end=54}{)}}{\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{1}} \htmlData{tutor-start=60,tutor-end=61}{=} \frac{\htmlData{tutor-start=68,tutor-end=69}{-}\htmlData{tutor-start=69,tutor-end=70}{1}\htmlData{tutor-start=70,tutor-end=71}{5}\htmlData{tutor-start=71,tutor-end=72}{a}_{\htmlData{tutor-start=74,tutor-end=75}{1}}}{\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=80}{1}} \htmlData{tutor-start=82,tutor-end=83}{=} \htmlData{tutor-start=84,tutor-end=85}{1}\htmlData{tutor-start=85,tutor-end=86}{5}\htmlData{tutor-start=86,tutor-end=87}{a}_{\htmlData{tutor-start=89,tutor-end=90}{1}}$。 或者直接累加：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{1}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{4}\htmlData{tutor-start=26,tutor-end=27}{a}_{\htmlData{tutor-start=29,tutor-end=30}{1}} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{8}\htmlData{tutor-start=35,tutor-end=36}{a}_{\htmlData{tutor-start=38,tutor-end=39}{1}} \htmlData{tutor-start=41,tutor-end=42}{=} \htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{5}\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{1}}$。

本步将上一步得到的表达式相除，得出最终数值。

详细展开： 1. 目标式：$\frac{\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=10}{4}}}{\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{2}}}$。 2. 代入：$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}}}{\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{1}}}$。 3. 消去 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$（因 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \neq \htmlData{tutor-start=11,tutor-end=12}{0}$）：$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{5}}{\htmlData{tutor-start=10,tutor-end=11}{2}}$。 4. 对比选项：(A) 2, (B) 4, (C) 15/2, (D) 17/2。 5. 结果与 (C) 一致。

本步分析程序框图的已知部分，理解变量 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 在当前流程中的含义。

详细展开： 1. 输入 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}$。 2. 第一个判断框通常是 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b}$ ?（或类似比较）。假设流程如下： - 先比较 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{b}$，将较大者赋值给 $\htmlData{tutor-start=0,tutor-end=1}{x}$。此时 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \max\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{)}$。 - 接下来需要将 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 与当前的最大值 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 进行比较，以确定三个数中的最大值。 3. 框图结构通常为： - $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a}$ - 如果 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}$，则 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{b}$ （此时 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}$ 中较大者） - 接下来进入空白判断框。 - 如果条件满足（“是”分支），执行 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{c}$。 - 如果条件不满足（“否”分支），保持 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 不变。 - 最后输出 $\htmlData{tutor-start=0,tutor-end=1}{x}$。 4. 既然目的是让 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 成为 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c}$ 中的最大值，且在“是”的分支里执行了 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}$，这意味着只有当 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 比当前的最大值 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 还要大时，才需要更新 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{c}$。

本步根据逻辑推导，从选项中选出正确的条件表达式。

详细展开： 1. 我们需要在 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}$ 时执行 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{c}$。 2. 观察选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{c}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{c}$ 3. 选项 (C) 和 (D) 仅涉及 $\htmlData{tutor-start=0,tutor-end=1}{b}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{c}$，忽略了 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的影响（已通过 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 体现），故排除。 4. 比较 (A) 和 (B)： - 若填 (A) $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}$：当 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 大于当前最大值 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 时，条件为真，执行 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}$，正确更新了最大值。 - 若填 (B) $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{c}$：当当前最大值 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 大于 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 时，条件为真，执行 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}$，这将用较小的 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 替换了较大的 $\htmlData{tutor-start=0,tutor-end=1}{x}$，逻辑错误。 5. 因此，应填入 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}$。

本步先忽略下标 $\htmlData{tutor-start=0,tutor-end=1}{i}$，解关于 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的不等式，找出 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的范围与参数 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的关系。

详细展开： 1. 原不等式：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{1}$。 2. 开平方去绝对值：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{|} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{1}$。 3. 等价于：$\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{1}$。 4. 各边减 1：$\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{x} \htmlData{tutor-start=9,tutor-end=10}{<} \htmlData{tutor-start=11,tutor-end=12}{0}$。 5. 各边乘以 -1（注意不等号方向改变）：$\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{2}$。 6. 因为题目已知 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{0}$，所以可以除以 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 而不改变不等号方向： $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \frac{\htmlData{tutor-start=14,tutor-end=15}{2}}{\htmlData{tutor-start=17,tutor-end=18}{a}}$。 7. 因此，对于每一个 $\htmlData{tutor-start=0,tutor-end=1}{i}$，不等式成立的充要条件是 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\left(\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{,} \frac{\htmlData{tutor-start=21,tutor-end=22}{2}}{\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{i}}}\right)$。

本步根据 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{3}}$ 的大小关系，确定区间端点 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{i}}}$ 的大小，从而求出公共部分。

详细展开： 1. 题目要求不等式对 $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{3}$ 都成立，即求三个解集的交集： $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \left(\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,} \frac{\htmlData{tutor-start=19,tutor-end=20}{2}}{\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{1}}}\right) \htmlData{tutor-start=36,tutor-end=41}{\cap }\left(\htmlData{tutor-start=47,tutor-end=48}{0}\htmlData{tutor-start=48,tutor-end=49}{,} \frac{\htmlData{tutor-start=56,tutor-end=57}{2}}{\htmlData{tutor-start=59,tutor-end=60}{a}_{\htmlData{tutor-start=62,tutor-end=63}{2}}}\right) \htmlData{tutor-start=73,tutor-end=78}{\cap }\left(\htmlData{tutor-start=84,tutor-end=85}{0}\htmlData{tutor-start=85,tutor-end=86}{,} \frac{\htmlData{tutor-start=93,tutor-end=94}{2}}{\htmlData{tutor-start=96,tutor-end=97}{a}_{\htmlData{tutor-start=99,tutor-end=100}{3}}}\right)$。 2. 已知 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{>} \htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{3}} \htmlData{tutor-start=22,tutor-end=23}{>} \htmlData{tutor-start=24,tutor-end=25}{0}$。 3. 取倒数并乘以 2，不等号方向改变： $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}} \htmlData{tutor-start=16,tutor-end=17}{<} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{2}}} \htmlData{tutor-start=34,tutor-end=35}{<} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{3}}}$ $\implies\htmlData{tutor-start=8,tutor-end=23}{ \frac{2}{a_{1}}} \htmlData{tutor-start=25,tutor-end=26}{<} \frac{\htmlData{tutor-start=33,tutor-end=34}{2}}{\htmlData{tutor-start=36,tutor-end=37}{a}_{\htmlData{tutor-start=39,tutor-end=40}{2}}} \htmlData{tutor-start=43,tutor-end=44}{<} \frac{\htmlData{tutor-start=51,tutor-end=52}{2}}{\htmlData{tutor-start=54,tutor-end=55}{a}_{\htmlData{tutor-start=57,tutor-end=58}{3}}}$。 4. 三个区间都是左端点为 0 的开区间，右端点越小，区间越短，包含于其他区间。 5. 最小的右端点是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}}$。 6. 因此，交集为 $\left(\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \frac{\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{1}}}\right)$。 7. 对比选项，(B) 符合。

观察分子分母中的角度，$70^\circ$ 与 $10^\circ$ 互余，且分母含有平方项，考虑利用诱导公式统一角度，并利用二倍角公式降幂。

详细展开： 首先处理分母中的 $\cos^{2} 10^\circ$。根据二倍角公式 $\cos \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=13}{\alpha }\htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{2}\cos^{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=32}{\alpha }\htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{1}$，可得 $\cos^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=16}{\alpha }\htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{+}\cos \htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=38}{\alpha}}{\htmlData{tutor-start=40,tutor-end=41}{2}}$。 所以，$2 - \cos^{2} 10^\circ = 2 - \frac{1 + \cos 20^\circ}{2} = \frac{4 - (1 + \cos 20^\circ)}{2} = \frac{3 - \cos 20^\circ}{2}$。 接下来处理分子中的 $\sin 70^\circ$。根据诱导公式 $\sin(90^\circ - \alpha) = \cos \alpha$，有 $\sin 70^\circ = \sin(90^\circ - 20^\circ) = \cos 20^\circ$。 因此，原式分子变为 $3 - \cos 20^\circ$。

将转化后的分子和分母代入原式，进行代数约分。

详细展开： 原式 $= \frac{3 - \cos 20^\circ}{\frac{3 - \cos 20^\circ}{2}}$。 由于 $\cos 20^\circ \neq 3$，分子 $3 - \cos 20^\circ$ 不为零，可以直接约分。 原式 $= (3 - \cos 20^\circ) \times \frac{2}{3 - \cos 20^\circ} = 2$。 故结果为 2，对应选项 (C)。

首先回顾向量共线（平行）的定义：方向相同或相反的非零向量叫平行向量，规定零向量与任意向量平行。我们需要检验选项是否覆盖了所有情况，特别是零向量的情形。

详细展开： 对于选项 (A)，若 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}$ 方向相反，它们依然共线，但方向不相同，故 (A) 是充分不必要条件。 对于选项 (B)，若 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,} \boldsymbol{\htmlData{tutor-start=34,tutor-end=35}{b}}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{,}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{)}$，两者均非零且共线，故“至少有一个为零向量”不是必要条件。 对于选项 (C)，表述为 $\htmlData{tutor-start=0,tutor-end=8}{\exists }\htmlData{tutor-start=8,tutor-end=16}{\lambda }\htmlData{tutor-start=16,tutor-end=20}{\in }\mathbf{\htmlData{tutor-start=28,tutor-end=29}{R}}\htmlData{tutor-start=30,tutor-end=31}{,} \boldsymbol{\htmlData{tutor-start=44,tutor-end=45}{b}}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=55}{\lambda }\boldsymbol{\htmlData{tutor-start=67,tutor-end=68}{a}}$。若 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\boldsymbol{\htmlData{tutor-start=27,tutor-end=28}{0}}$ 且 $\boldsymbol{b} \neq \boldsymbol{0}$，此时两向量共线（因为零向量与任意向量共线），但不存在实数 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}$ 使得 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=23}{\lambda }\boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{0}}\htmlData{tutor-start=37,tutor-end=38}{=}\boldsymbol{\htmlData{tutor-start=50,tutor-end=51}{0}}$。因此该条件漏掉了 $\boldsymbol{a}=\boldsymbol{0}, \boldsymbol{b} \neq \boldsymbol{0}$ 的情况，不是必要条件。

分析选项 (D)。这是向量共线（线性相关）的标准代数定义。

详细展开： 充分性：若存在不全为零的实数 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=20}{\lambda}_{\htmlData{tutor-start=22,tutor-end=23}{2}}$ 使得 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}} \boldsymbol{\htmlData{tutor-start=24,tutor-end=25}{a}} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=36}{\lambda}_{\htmlData{tutor-start=38,tutor-end=39}{2}} \boldsymbol{\htmlData{tutor-start=53,tutor-end=54}{b}} \htmlData{tutor-start=56,tutor-end=57}{=} \boldsymbol{\htmlData{tutor-start=70,tutor-end=71}{0}}$。 若 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}} \neq \htmlData{tutor-start=17,tutor-end=18}{0}$，则 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{-}\frac{\htmlData{tutor-start=24,tutor-end=31}{\lambda}_{\htmlData{tutor-start=33,tutor-end=34}{2}}}{\htmlData{tutor-start=37,tutor-end=44}{\lambda}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \boldsymbol{\htmlData{tutor-start=62,tutor-end=63}{b}}$，即 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 与 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}$ 共线； 若 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0}$，则必有 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \neq \htmlData{tutor-start=17,tutor-end=18}{0}$，从而 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} \htmlData{tutor-start=15,tutor-end=16}{=} \boldsymbol{\htmlData{tutor-start=29,tutor-end=30}{0}}$，零向量与 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 共线。 必要性：若 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}$ 共线。 若 $\boldsymbol{a} \neq \boldsymbol{0}$，则存在 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}$ 使 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=25}{\lambda }\boldsymbol{\htmlData{tutor-start=37,tutor-end=38}{a}}$，即 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{a}} \htmlData{tutor-start=23,tutor-end=24}{-} \boldsymbol{\htmlData{tutor-start=37,tutor-end=38}{b}} \htmlData{tutor-start=40,tutor-end=41}{=} \boldsymbol{\htmlData{tutor-start=54,tutor-end=55}{0}}$，取 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=19}{\lambda}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=28}{\lambda}_{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{1}$（不全为零）； 若 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{=} \boldsymbol{\htmlData{tutor-start=29,tutor-end=30}{0}}$，取 $\htmlData{tutor-start=0,tutor-end=7}{\lambda}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=22}{\lambda}_{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}$，则 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=7}{\cdot }\boldsymbol{\htmlData{tutor-start=19,tutor-end=20}{0}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=31}{\cdot }\boldsymbol{\htmlData{tutor-start=43,tutor-end=44}{b}} \htmlData{tutor-start=46,tutor-end=47}{=} \boldsymbol{\htmlData{tutor-start=60,tutor-end=61}{0}}$，满足条件。 综上，(D) 是充要条件。

首先从5天中选出3天作为志愿者活动日，然后在这3天中安排甲、乙、丙三人，需满足甲的时间早于乙和丙。

详细展开： 第一步：从周一至周五的 5 天中任选 3 天。组合数为 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{3}}$。 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{5} \htmlData{tutor-start=20,tutor-end=27}{\times }\htmlData{tutor-start=27,tutor-end=28}{4} \htmlData{tutor-start=29,tutor-end=36}{\times }\htmlData{tutor-start=36,tutor-end=37}{3}}{\htmlData{tutor-start=39,tutor-end=40}{3} \htmlData{tutor-start=41,tutor-end=48}{\times }\htmlData{tutor-start=48,tutor-end=49}{2} \htmlData{tutor-start=50,tutor-end=57}{\times }\htmlData{tutor-start=57,tutor-end=58}{1}} \htmlData{tutor-start=60,tutor-end=61}{=} \htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{0}$ 种选法。 第二步：在选定的 3 天中安排甲、乙、丙。设这 3 天按时间先后为第 1 天、第 2 天、第 3 天。 题目要求“甲安排在另外两位前面”，这意味着甲必须占据这 3 天中最早的那一天（即第 1 天）。 一旦甲的位置确定（只能是第 1 天），剩下的乙和丙可以在剩下的 2 天（第 2 天、第 3 天）中任意排列。 乙和丙的排列数为 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2}$ 种（乙前丙后，或丙前乙后）。 根据分步乘法计数原理，总方法数为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=10}{\times }\htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=19}{\times }\htmlData{tutor-start=19,tutor-end=20}{2} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{0}$ 种。

利用对称性原理快速验证结果。

详细展开： 若不考虑顺序限制，从 5 天中选 3 天并安排甲乙丙三人的总排列数为 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{5} \htmlData{tutor-start=14,tutor-end=21}{\times }\htmlData{tutor-start=21,tutor-end=22}{4} \htmlData{tutor-start=23,tutor-end=30}{\times }\htmlData{tutor-start=30,tutor-end=31}{3} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{6}\htmlData{tutor-start=35,tutor-end=36}{0}$ 种。 在任意选定的 3 天中，甲、乙、丙三人的相对位置共有 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{!} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{6}$ 种可能。 其中，满足“甲在乙、丙前面”的情况，即甲排在第一位，乙、丙在后两位。这样的排列有 2 种（甲乙丙、甲丙乙）。 或者更简单地看，甲、乙、丙三人谁排在最前面是等可能的。甲排在最前面的概率是 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$。 但是题目说的是“甲在另外两位前面”，也就是甲是三人中最早的。在三人的所有 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{!}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{6}$ 种相对排序中，甲排第一的情况有 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{!} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}$ 种（甲乙丙，甲丙乙）。 所以符合条件的排列数是总排列数的 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{6}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{3}}$。 $\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=10}{\times }\frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{3}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{0}$ 种。 结果一致。

分析题目给出的边界条件，确定被积函数和积分区间。由于 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}}$ 在区间 $\htmlData{tutor-start=0,tutor-end=1}{[}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{]}$ 上恒大于 0，且 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴即 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$，故所求面积即为函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}}$ 在该区间上的定积分。

详细展开： 1. 确定积分下限 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}$ 和上限 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}$。 2. 确定被积函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{x}}$。 3. 根据定积分的几何意义，面积 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \int_{\frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{2}}}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \frac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{x}} \htmlData{tutor-start=39,tutor-end=40}{d}\htmlData{tutor-start=40,tutor-end=41}{x}$。

利用牛顿-莱布尼茨公式计算定积分的值，并结合对数运算性质化简结果。

详细展开： 1. 找到原函数：$\htmlData{tutor-start=0,tutor-end=1}{(}\ln \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)}' \htmlData{tutor-start=9,tutor-end=10}{=} \frac{\htmlData{tutor-start=17,tutor-end=18}{1}}{\htmlData{tutor-start=20,tutor-end=21}{x}}$，所以 $\int \frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{x}} \htmlData{tutor-start=17,tutor-end=18}{d}\htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{=} \ln\htmlData{tutor-start=25,tutor-end=26}{|}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{|} \htmlData{tutor-start=29,tutor-end=30}{+} \htmlData{tutor-start=31,tutor-end=32}{C}$。 2. 代入上下限：$\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{[}\ln \htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{]}_{\frac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\htmlData{tutor-start=22,tutor-end=23}{2}}}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{=} \ln \htmlData{tutor-start=36,tutor-end=37}{2} \htmlData{tutor-start=38,tutor-end=39}{-} \ln\htmlData{tutor-start=43,tutor-end=44}{(}\frac{\htmlData{tutor-start=50,tutor-end=51}{1}}{\htmlData{tutor-start=53,tutor-end=54}{2}}\htmlData{tutor-start=55,tutor-end=56}{)}$。 3. 化简：$\ln\htmlData{tutor-start=3,tutor-end=4}{(}\frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{=} \ln\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{2}^{\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{)} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{-}\ln \htmlData{tutor-start=38,tutor-end=39}{2}$。 4. 最终结果：$\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \ln \htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\ln \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2}\ln \htmlData{tutor-start=28,tutor-end=29}{2}$。 5. 对比选项，(D) 符合。

首先确定抛物线的焦点和准线方程。根据抛物线定义，抛物线上任意一点到焦点的距离等于该点到准线的距离。因此，题目要求的“点 P 到 Q 的距离 + 点 P 到焦点 F 的距离”可以转化为“点 P 到 Q 的距离 + 点 P 到准线 l 的距离”。

详细展开： 1. 抛物线方程 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{x}$，标准形式 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{p}\htmlData{tutor-start=10,tutor-end=11}{x}$，故 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4} \htmlData{tutor-start=5,tutor-end=17}{\Rightarrow }\htmlData{tutor-start=17,tutor-end=18}{p}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{2}$。 2. 焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，准线方程 $\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}$。 3. 设 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{0}}\htmlData{tutor-start=14,tutor-end=15}{)}$，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{|}$ 等于 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 到直线 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}$ 的距离，记为 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{l}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{x}_{\htmlData{tutor-start=28,tutor-end=29}{0}} \htmlData{tutor-start=31,tutor-end=32}{+} \htmlData{tutor-start=33,tutor-end=34}{1}$。 4. 目标是最小化 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{+} \htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{l}\htmlData{tutor-start=13,tutor-end=14}{)}$。

几何上，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{+} \htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{l}\htmlData{tutor-start=13,tutor-end=14}{)}$ 表示点 Q 到点 P 的距离加上点 P 到准线的水平距离。要使这个和最小，点 P 应该位于过点 Q 且垂直于准线的直线上。因为两点之间线段最短，且垂线段最短，当 Q、P 以及 P 在准线上的投影点共线且垂直于准线时，距离之和最小。

详细展开： 1. 准线是竖直直线 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}$，所以垂直于准线的线是水平线（纵坐标恒定）。 2. 点 $\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的纵坐标为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$。 3. 因此，取得最小值时，点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 的纵坐标 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{P}}$ 必须等于 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$。 4. 将 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{P}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}$ 代入抛物线方程 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{x}$： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=26}{\Rightarrow }\htmlData{tutor-start=26,tutor-end=27}{1} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{4}\htmlData{tutor-start=31,tutor-end=32}{x} \htmlData{tutor-start=33,tutor-end=45}{\Rightarrow }\htmlData{tutor-start=45,tutor-end=46}{x} \htmlData{tutor-start=47,tutor-end=48}{=} \frac{\htmlData{tutor-start=55,tutor-end=56}{1}}{\htmlData{tutor-start=58,tutor-end=59}{4}}$。 5. 此时点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 的坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}$。 6. 验证：点 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}$ 确实在抛物线上。此时 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{+} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{|}$ 最小。 7. 对比选项，(A) 符合。

将该棱视为长方体的一条对角线，或者直接在空间直角坐标系中设该棱对应的向量为 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{v}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{y}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{z}\htmlData{tutor-start=25,tutor-end=26}{)}$。 设棱长为 $\htmlData{tutor-start=0,tutor-end=1}{L} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{7}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{z}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{7}$。

详细展开： 正视图投影长为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{6}}$。正视图通常是在 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{z}$ 平面或 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{z}$ 平面上的投影，取决于坐标系的建立习惯。在高中数学常规设定中： 正视图（主视图）：投影到 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{z}$ 平面（或垂直于 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴的平面），投影长平方为 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{z}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{(}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{6}}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{6}$。 侧视图（左视图）：投影到 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{z}$ 平面（或垂直于 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴的平面），投影长为 $\htmlData{tutor-start=0,tutor-end=1}{a}$，则 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{z}^{\htmlData{tutor-start=19,tutor-end=20}{2}}$。 俯视图：投影到 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y}$ 平面（或垂直于 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 轴的平面），投影长为 $\htmlData{tutor-start=0,tutor-end=1}{b}$，则 $\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{y}^{\htmlData{tutor-start=19,tutor-end=20}{2}}$。

由 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{z}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{7}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{z}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{6}$，可得： $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{7} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{6} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1}$。 进而： $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{z}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{1} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{z}^{\htmlData{tutor-start=31,tutor-end=32}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{y}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{1}$ 又因为 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{z}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{6}$，我们可以设 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{u}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{z}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{v}$，则 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=11}{\ge}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{v}\htmlData{tutor-start=15,tutor-end=18}{\ge}\htmlData{tutor-start=18,tutor-end=19}{0}$。 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{v}}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{b} \htmlData{tutor-start=18,tutor-end=19}{=} \sqrt{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{u}}$。 我们需要求 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b} \htmlData{tutor-start=4,tutor-end=5}{=} \sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{v}} \htmlData{tutor-start=17,tutor-end=18}{+} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{u}}$ 的最大值。

目标函数 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{v}} \htmlData{tutor-start=15,tutor-end=16}{+} \sqrt{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{u}}$，约束条件 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{v} \htmlData{tutor-start=11,tutor-end=15}{\ge }\htmlData{tutor-start=15,tutor-end=16}{0}$。

详细展开： 方法一：柯西不等式 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=8}{\cdot }\sqrt{\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{v}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=28}{\cdot }\sqrt{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{u}}\htmlData{tutor-start=38,tutor-end=39}{)}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=48}{\le }\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{1}^{\htmlData{tutor-start=52,tutor-end=53}{2}}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{1}^{\htmlData{tutor-start=58,tutor-end=59}{2}}\htmlData{tutor-start=60,tutor-end=61}{)}\htmlData{tutor-start=61,tutor-end=62}{(} \htmlData{tutor-start=63,tutor-end=64}{(}\htmlData{tutor-start=64,tutor-end=65}{1}\htmlData{tutor-start=65,tutor-end=66}{+}\htmlData{tutor-start=66,tutor-end=67}{v}\htmlData{tutor-start=67,tutor-end=68}{)} \htmlData{tutor-start=69,tutor-end=70}{+} \htmlData{tutor-start=71,tutor-end=72}{(}\htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{+}\htmlData{tutor-start=74,tutor-end=75}{u}\htmlData{tutor-start=75,tutor-end=76}{)} \htmlData{tutor-start=77,tutor-end=78}{)} \htmlData{tutor-start=79,tutor-end=80}{=} \htmlData{tutor-start=81,tutor-end=82}{2}\htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{2}\htmlData{tutor-start=84,tutor-end=85}{+}\htmlData{tutor-start=85,tutor-end=86}{u}\htmlData{tutor-start=86,tutor-end=87}{+}\htmlData{tutor-start=87,tutor-end=88}{v}\htmlData{tutor-start=88,tutor-end=89}{)} \htmlData{tutor-start=90,tutor-end=91}{=} \htmlData{tutor-start=92,tutor-end=93}{2}\htmlData{tutor-start=93,tutor-end=94}{(}\htmlData{tutor-start=94,tutor-end=95}{2}\htmlData{tutor-start=95,tutor-end=96}{+}\htmlData{tutor-start=96,tutor-end=97}{6}\htmlData{tutor-start=97,tutor-end=98}{)} \htmlData{tutor-start=99,tutor-end=100}{=} \htmlData{tutor-start=101,tutor-end=102}{1}\htmlData{tutor-start=102,tutor-end=103}{6}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=6}{\le }\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{6}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{4}$。 当且仅当 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}} \htmlData{tutor-start=11,tutor-end=12}{=} \sqrt{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{u}}$，即 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{v}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}$ 时取等号。 此时 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{z}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}$，满足 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{z}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{6}$。

方法二：函数法 令 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{u}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{6}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{u}\htmlData{tutor-start=19,tutor-end=20}{)}} \htmlData{tutor-start=22,tutor-end=23}{+} \sqrt{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{u}} \htmlData{tutor-start=35,tutor-end=36}{=} \sqrt{\htmlData{tutor-start=43,tutor-end=44}{7}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{u}} \htmlData{tutor-start=48,tutor-end=49}{+} \sqrt{\htmlData{tutor-start=56,tutor-end=57}{1}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{u}}$，其中 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{u} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{6}$。 求导 $\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{7}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{u}}} \htmlData{tutor-start=31,tutor-end=32}{+} \frac{\htmlData{tutor-start=39,tutor-end=40}{1}}{\htmlData{tutor-start=42,tutor-end=43}{2}\sqrt{\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{+}\htmlData{tutor-start=51,tutor-end=52}{u}}}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0} \htmlData{tutor-start=8,tutor-end=20}{\Rightarrow }\sqrt{\htmlData{tutor-start=26,tutor-end=27}{7}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{u}} \htmlData{tutor-start=31,tutor-end=32}{=} \sqrt{\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{u}} \htmlData{tutor-start=44,tutor-end=56}{\Rightarrow }\htmlData{tutor-start=56,tutor-end=57}{7}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{u}\htmlData{tutor-start=59,tutor-end=60}{=}\htmlData{tutor-start=60,tutor-end=61}{1}\htmlData{tutor-start=61,tutor-end=62}{+}\htmlData{tutor-start=62,tutor-end=63}{u} \htmlData{tutor-start=64,tutor-end=76}{\Rightarrow }\htmlData{tutor-start=76,tutor-end=77}{2}\htmlData{tutor-start=77,tutor-end=78}{u}\htmlData{tutor-start=78,tutor-end=79}{=}\htmlData{tutor-start=79,tutor-end=80}{6} \htmlData{tutor-start=81,tutor-end=93}{\Rightarrow }\htmlData{tutor-start=93,tutor-end=94}{u}\htmlData{tutor-start=94,tutor-end=95}{=}\htmlData{tutor-start=95,tutor-end=96}{3}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 时，$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{4}} \htmlData{tutor-start=16,tutor-end=17}{+} \sqrt{\htmlData{tutor-start=24,tutor-end=25}{4}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{4}$。 端点值：$\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} \htmlData{tutor-start=4,tutor-end=16}{\Rightarrow }\htmlData{tutor-start=16,tutor-end=17}{f}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{=}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{7}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=40}{\approx }\htmlData{tutor-start=40,tutor-end=41}{3}\htmlData{tutor-start=41,tutor-end=42}{.}\htmlData{tutor-start=42,tutor-end=43}{6}\htmlData{tutor-start=43,tutor-end=44}{5} \htmlData{tutor-start=45,tutor-end=46}{<} \htmlData{tutor-start=47,tutor-end=48}{4}$；$\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6} \htmlData{tutor-start=4,tutor-end=16}{\Rightarrow }\htmlData{tutor-start=16,tutor-end=17}{f}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{6}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{+}\sqrt{\htmlData{tutor-start=29,tutor-end=30}{7}} \htmlData{tutor-start=32,tutor-end=33}{<} \htmlData{tutor-start=34,tutor-end=35}{4}$。 故最大值为 4。

对应选项 (C)。

首先根据向量数乘和加法的坐标运算规则，求出向量 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{a}} \htmlData{tutor-start=23,tutor-end=24}{+} \boldsymbol{\htmlData{tutor-start=37,tutor-end=38}{b}}$ 的坐标表达式。

详细展开： 1. 已知 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}$，$\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{)}$。 2. 计算 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{a}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=37}{\lambda}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=46}{\lambda}\htmlData{tutor-start=46,tutor-end=47}{)}$。 3. 计算 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\boldsymbol{\htmlData{tutor-start=20,tutor-end=21}{a}} \htmlData{tutor-start=23,tutor-end=24}{+} \boldsymbol{\htmlData{tutor-start=37,tutor-end=38}{b}} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{0}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{4}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=56}{\lambda}\htmlData{tutor-start=56,tutor-end=57}{+}\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{,} \htmlData{tutor-start=60,tutor-end=67}{\lambda}\htmlData{tutor-start=67,tutor-end=68}{+}\htmlData{tutor-start=68,tutor-end=69}{0}\htmlData{tutor-start=69,tutor-end=70}{)} \htmlData{tutor-start=71,tutor-end=72}{=} \htmlData{tutor-start=73,tutor-end=74}{(}\htmlData{tutor-start=74,tutor-end=75}{4}\htmlData{tutor-start=75,tutor-end=76}{,} \htmlData{tutor-start=77,tutor-end=78}{1}\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=86}{\lambda}\htmlData{tutor-start=86,tutor-end=87}{,} \htmlData{tutor-start=88,tutor-end=95}{\lambda}\htmlData{tutor-start=95,tutor-end=96}{)}$。

利用向量模长公式 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{v}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{y}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{z}^{\htmlData{tutor-start=40,tutor-end=41}{2}}}$ 建立方程，结合 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{>} \htmlData{tutor-start=10,tutor-end=11}{0}$ 的条件求解。

详细展开： 1. 模长平方：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=9}{\lambda }\boldsymbol{\htmlData{tutor-start=21,tutor-end=22}{a}} \htmlData{tutor-start=24,tutor-end=25}{+} \boldsymbol{\htmlData{tutor-start=38,tutor-end=39}{b}}\htmlData{tutor-start=40,tutor-end=41}{|}^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{=} \htmlData{tutor-start=48,tutor-end=49}{4}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{+} \htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=66}{\lambda}\htmlData{tutor-start=66,tutor-end=67}{)}^{\htmlData{tutor-start=69,tutor-end=70}{2}} \htmlData{tutor-start=72,tutor-end=73}{+} \htmlData{tutor-start=74,tutor-end=81}{\lambda}^{\htmlData{tutor-start=83,tutor-end=84}{2}}$。 2. 题目已知模长为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{9}}$，故模长平方为 29。 3. 建立方程：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=19}{\lambda }\htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=28}{\lambda}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{)} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=43}{\lambda}^{\htmlData{tutor-start=45,tutor-end=46}{2}} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{9}$。 4. 整理得：$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=8}{\lambda}^{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=24}{\lambda }\htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{7} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{9} \htmlData{tutor-start=34,tutor-end=46}{\Rightarrow }\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=54}{\lambda}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{-} \htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=70}{\lambda }\htmlData{tutor-start=70,tutor-end=71}{-} \htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{2} \htmlData{tutor-start=75,tutor-end=76}{=} \htmlData{tutor-start=77,tutor-end=78}{0}$。 5. 化简：$\htmlData{tutor-start=0,tutor-end=7}{\lambda}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=22}{\lambda }\htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{6} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{0}$。 6. 因式分解：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=9}{\lambda }\htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=22}{\lambda }\htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{)} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{0}$。 7. 解得 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{3}$ 或 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}$。 8. 根据条件 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{>} \htmlData{tutor-start=10,tutor-end=11}{0}$，舍去 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}$，故 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{3}$。

从双曲线标准方程中提取 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}$ 的值，确定顶点 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 的坐标，以及渐近线斜率。

详细展开： 1. 双曲线方程 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{y}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{6}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1}$。 2. 参数：$\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{9} \htmlData{tutor-start=8,tutor-end=20}{\Rightarrow }\htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{3}$，$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{6} \htmlData{tutor-start=9,tutor-end=21}{\Rightarrow }\htmlData{tutor-start=21,tutor-end=22}{b}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{4}$。 3. 半焦距：$\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} \htmlData{tutor-start=23,tutor-end=24}{=} \sqrt{\htmlData{tutor-start=31,tutor-end=32}{9}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{6}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{5}$。 4. 右顶点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，右焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$。 5. 渐近线方程：$\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{b}}{\htmlData{tutor-start=17,tutor-end=18}{a}}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=27}{\pm }\frac{\htmlData{tutor-start=33,tutor-end=34}{4}}{\htmlData{tutor-start=36,tutor-end=37}{3}}\htmlData{tutor-start=38,tutor-end=39}{x}$。斜率为 $\htmlData{tutor-start=0,tutor-end=4}{\pm }\frac{\htmlData{tutor-start=10,tutor-end=11}{4}}{\htmlData{tutor-start=13,tutor-end=14}{3}}$。

写出过点 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 且平行于一条渐近线的直线方程，并联立双曲线方程求出交点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的坐标。

详细展开： 1. 不妨取斜率为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ 的渐近线（由对称性，取另一条结果面积相同）。 2. 直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{F}$ 过点 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，斜率 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{4}}{\htmlData{tutor-start=13,tutor-end=14}{3}}$。 3. 直线方程：$\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{0} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=26}{)} \htmlData{tutor-start=27,tutor-end=39}{\Rightarrow }\htmlData{tutor-start=39,tutor-end=40}{y} \htmlData{tutor-start=41,tutor-end=42}{=} \frac{\htmlData{tutor-start=49,tutor-end=50}{4}}{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{x} \htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=60}{5}\htmlData{tutor-start=60,tutor-end=61}{)}$。 4. 联立双曲线方程： $\begin{cases} \frac{\htmlData{tutor-start=20,tutor-end=21}{x}^{\htmlData{tutor-start=23,tutor-end=24}{2}}}{\htmlData{tutor-start=27,tutor-end=28}{9}} \htmlData{tutor-start=30,tutor-end=31}{-} \frac{\htmlData{tutor-start=38,tutor-end=39}{y}^{\htmlData{tutor-start=41,tutor-end=42}{2}}}{\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{6}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{1} \\ \htmlData{tutor-start=56,tutor-end=57}{y} \htmlData{tutor-start=58,tutor-end=59}{=} \frac{\htmlData{tutor-start=66,tutor-end=67}{4}}{\htmlData{tutor-start=69,tutor-end=70}{3}}\htmlData{tutor-start=71,tutor-end=72}{(}\htmlData{tutor-start=72,tutor-end=73}{x} \htmlData{tutor-start=74,tutor-end=75}{-} \htmlData{tutor-start=76,tutor-end=77}{5}\htmlData{tutor-start=77,tutor-end=78}{)} \end{cases}$ 5. 代入消元：$\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{[}\frac{\htmlData{tutor-start=31,tutor-end=32}{4}}{\htmlData{tutor-start=34,tutor-end=35}{3}}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{x}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{]}^{\htmlData{tutor-start=44,tutor-end=45}{2}}}{\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{6}} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=55}{1}$。 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\frac{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{6}}{\htmlData{tutor-start=34,tutor-end=35}{9}}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{x}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{)}^{\htmlData{tutor-start=43,tutor-end=44}{2}}}{\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{6}} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{1}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{5}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{9}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{1}$ $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{9}$ $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=26}{)} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{9}$ $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=5}{-} \htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{5} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{9} \htmlData{tutor-start=13,tutor-end=25}{\Rightarrow }\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{x} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{4} \htmlData{tutor-start=34,tutor-end=46}{\Rightarrow }\htmlData{tutor-start=46,tutor-end=47}{x} \htmlData{tutor-start=48,tutor-end=49}{=} \frac{\htmlData{tutor-start=56,tutor-end=57}{1}\htmlData{tutor-start=57,tutor-end=58}{7}}{\htmlData{tutor-start=60,tutor-end=61}{5}}$。 6. 求 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{B}}$：$\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{B}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{(}\frac{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{7}}{\htmlData{tutor-start=30,tutor-end=31}{5}} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{5}\htmlData{tutor-start=36,tutor-end=37}{)} \htmlData{tutor-start=38,tutor-end=39}{=} \frac{\htmlData{tutor-start=46,tutor-end=47}{4}}{\htmlData{tutor-start=49,tutor-end=50}{3}}\htmlData{tutor-start=51,tutor-end=52}{(}\frac{\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{7}\htmlData{tutor-start=60,tutor-end=61}{-}\htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=63}{5}}{\htmlData{tutor-start=65,tutor-end=66}{5}}\htmlData{tutor-start=67,tutor-end=68}{)} \htmlData{tutor-start=69,tutor-end=70}{=} \frac{\htmlData{tutor-start=77,tutor-end=78}{4}}{\htmlData{tutor-start=80,tutor-end=81}{3}}\htmlData{tutor-start=82,tutor-end=83}{(}\frac{\htmlData{tutor-start=89,tutor-end=90}{-}\htmlData{tutor-start=90,tutor-end=91}{8}}{\htmlData{tutor-start=93,tutor-end=94}{5}}\htmlData{tutor-start=95,tutor-end=96}{)} \htmlData{tutor-start=97,tutor-end=98}{=} \htmlData{tutor-start=99,tutor-end=100}{-}\frac{\htmlData{tutor-start=106,tutor-end=107}{3}\htmlData{tutor-start=107,tutor-end=108}{2}}{\htmlData{tutor-start=110,tutor-end=111}{1}\htmlData{tutor-start=111,tutor-end=112}{5}}$。 7. 故点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{7}}{\htmlData{tutor-start=11,tutor-end=12}{5}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{-}\frac{\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{2}}{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{5}}\htmlData{tutor-start=29,tutor-end=30}{)}$。