2009 年高考数学（大纲卷 2理科）

首先处理集合 B 中的不等式条件。分式不等式通常转化为整式不等式组或符号法则求解，需注意分母不为零的限制。

详细展开： 集合 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\left\{\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=16}{\mid }\frac{\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{4}}\htmlData{tutor-start=31,tutor-end=32}{<}\htmlData{tutor-start=32,tutor-end=33}{0}\right\}$。 由分式小于 0 可知，分子与分母异号。 即 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{0}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4} \neq \htmlData{tutor-start=9,tutor-end=10}{0}$。 二次函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{)}$ 开口向上，零点为 $\htmlData{tutor-start=0,tutor-end=1}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{4}$。 小于 0 的部分位于两根之间，即 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{4}$。 因此，集合 $\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{)}$。

在明确集合 A 和集合 B 的区间表示后，寻找它们的公共部分即为交集。

详细展开： 已知集合 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{>}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=16}{\}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=30}{\infty}\htmlData{tutor-start=30,tutor-end=31}{)}$。 已求得集合 $\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{)}$。 求 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B}$，即寻找同时满足 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{3}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{4}$ 的 $\htmlData{tutor-start=0,tutor-end=1}{x}$。 在数轴上表示： $\htmlData{tutor-start=0,tutor-end=1}{A}$: 从 3 向右延伸（不含 3）。 $\htmlData{tutor-start=0,tutor-end=1}{B}$: 从 1 到 4 之间（不含 1 和 4）。 公共部分为 $\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{4}$。 用区间表示为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{)}$。 对比选项，(B) 符合。

根据 $\cot \htmlData{tutor-start=5,tutor-end=6}{A}$ 的符号判断角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 所在的象限，从而确定 $\cos \htmlData{tutor-start=5,tutor-end=6}{A}$ 的符号。

详细展开： 已知 $\cot \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}}{\htmlData{tutor-start=20,tutor-end=21}{5}} \htmlData{tutor-start=23,tutor-end=24}{<} \htmlData{tutor-start=25,tutor-end=26}{0}$。 在三角形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 中，$\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=13}{\pi}\htmlData{tutor-start=13,tutor-end=14}{)}$。 余切值为负，说明 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是第二象限角（钝角）。 在第二象限，$\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0}$，$\sin \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{>} \htmlData{tutor-start=9,tutor-end=10}{0}$。 因此，最终结果 $\cos \htmlData{tutor-start=5,tutor-end=6}{A}$ 必为负值。排除选项 (A) 和 (B)。

利用 $\cot \htmlData{tutor-start=5,tutor-end=6}{A}$ 与 $\cos \htmlData{tutor-start=5,tutor-end=6}{A}$ 的关系或构造直角三角形模型计算绝对值，再加符号。

详细展开： 方法一：公式法。 $\csc^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{+} \cot^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{A} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{-}\frac{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}}{\htmlData{tutor-start=46,tutor-end=47}{5}}\htmlData{tutor-start=48,tutor-end=49}{)}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{1} \htmlData{tutor-start=58,tutor-end=59}{+} \frac{\htmlData{tutor-start=66,tutor-end=67}{1}\htmlData{tutor-start=67,tutor-end=68}{4}\htmlData{tutor-start=68,tutor-end=69}{4}}{\htmlData{tutor-start=71,tutor-end=72}{2}\htmlData{tutor-start=72,tutor-end=73}{5}} \htmlData{tutor-start=75,tutor-end=76}{=} \frac{\htmlData{tutor-start=83,tutor-end=84}{1}\htmlData{tutor-start=84,tutor-end=85}{6}\htmlData{tutor-start=85,tutor-end=86}{9}}{\htmlData{tutor-start=88,tutor-end=89}{2}\htmlData{tutor-start=89,tutor-end=90}{5}}$。 $\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \frac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\csc^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{A}} \htmlData{tutor-start=34,tutor-end=35}{=} \frac{\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{5}}{\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{6}\htmlData{tutor-start=48,tutor-end=49}{9}}$。 $\cos^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{-} \sin^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{A} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{1} \htmlData{tutor-start=32,tutor-end=33}{-} \frac{\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{5}}{\htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{6}\htmlData{tutor-start=46,tutor-end=47}{9}} \htmlData{tutor-start=49,tutor-end=50}{=} \frac{\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{4}\htmlData{tutor-start=59,tutor-end=60}{4}}{\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{6}\htmlData{tutor-start=64,tutor-end=65}{9}}$。 $\htmlData{tutor-start=0,tutor-end=1}{|}\cos \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{=} \sqrt{\frac{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{4}}{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{6}\htmlData{tutor-start=30,tutor-end=31}{9}}} \htmlData{tutor-start=34,tutor-end=35}{=} \frac{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}}{\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{3}}$。 因 $\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0}$，故 $\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}}{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{3}}$。

方法二：定义法。 设角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 终边上一点 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{y}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}$。 $\cot \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{x}}{\htmlData{tutor-start=18,tutor-end=19}{y}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{-}\frac{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{2}}{\htmlData{tutor-start=34,tutor-end=35}{5}}$。因 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 在第二象限，取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{5}$。 $\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{5}^{\htmlData{tutor-start=23,tutor-end=24}{2}}} \htmlData{tutor-start=27,tutor-end=28}{=} \sqrt{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{4}\htmlData{tutor-start=37,tutor-end=38}{4}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{5}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{3}$。 $\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{x}}{\htmlData{tutor-start=18,tutor-end=19}{r}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{-}\frac{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{2}}{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{3}}$。 对比选项，(D) 符合。

根据导数的几何意义，曲线在某点的切线斜率等于该点处的导数值。需先求导函数，再代入横坐标。

详细展开： 函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{x}}{\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}}$。 使用商的求导法则：$\left(\frac{\htmlData{tutor-start=12,tutor-end=13}{u}}{\htmlData{tutor-start=15,tutor-end=16}{v}}\right)' \htmlData{tutor-start=26,tutor-end=27}{=} \frac{\htmlData{tutor-start=34,tutor-end=35}{u}'\htmlData{tutor-start=36,tutor-end=37}{v} \htmlData{tutor-start=38,tutor-end=39}{-} \htmlData{tutor-start=40,tutor-end=41}{u}\htmlData{tutor-start=41,tutor-end=42}{v}'}{\htmlData{tutor-start=45,tutor-end=46}{v}^{\htmlData{tutor-start=48,tutor-end=49}{2}}}$。 令 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{v}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{u}'\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{v}'\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}$。 $\htmlData{tutor-start=0,tutor-end=1}{y}' \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1} \htmlData{tutor-start=13,tutor-end=19}{\cdot }\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{x} \htmlData{tutor-start=30,tutor-end=36}{\cdot }\htmlData{tutor-start=36,tutor-end=37}{2}}{\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{x}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{)}^{\htmlData{tutor-start=47,tutor-end=48}{2}}} \htmlData{tutor-start=51,tutor-end=52}{=} \frac{\htmlData{tutor-start=59,tutor-end=60}{2}\htmlData{tutor-start=60,tutor-end=61}{x} \htmlData{tutor-start=62,tutor-end=63}{-} \htmlData{tutor-start=64,tutor-end=65}{1} \htmlData{tutor-start=66,tutor-end=67}{-} \htmlData{tutor-start=68,tutor-end=69}{2}\htmlData{tutor-start=69,tutor-end=70}{x}}{\htmlData{tutor-start=72,tutor-end=73}{(}\htmlData{tutor-start=73,tutor-end=74}{2}\htmlData{tutor-start=74,tutor-end=75}{x}\htmlData{tutor-start=75,tutor-end=76}{-}\htmlData{tutor-start=76,tutor-end=77}{1}\htmlData{tutor-start=77,tutor-end=78}{)}^{\htmlData{tutor-start=80,tutor-end=81}{2}}} \htmlData{tutor-start=84,tutor-end=85}{=} \frac{\htmlData{tutor-start=92,tutor-end=93}{-}\htmlData{tutor-start=93,tutor-end=94}{1}}{\htmlData{tutor-start=96,tutor-end=97}{(}\htmlData{tutor-start=97,tutor-end=98}{2}\htmlData{tutor-start=98,tutor-end=99}{x}\htmlData{tutor-start=99,tutor-end=100}{-}\htmlData{tutor-start=100,tutor-end=101}{1}\htmlData{tutor-start=101,tutor-end=102}{)}^{\htmlData{tutor-start=104,tutor-end=105}{2}}}$。 切点为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$，即 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$。 代入 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 计算斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}$： $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{y}'\htmlData{tutor-start=6,tutor-end=7}{|}_{\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=34}{\cdot }\htmlData{tutor-start=34,tutor-end=35}{1} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{)}^{\htmlData{tutor-start=42,tutor-end=43}{2}}} \htmlData{tutor-start=46,tutor-end=47}{=} \frac{\htmlData{tutor-start=54,tutor-end=55}{-}\htmlData{tutor-start=55,tutor-end=56}{1}}{\htmlData{tutor-start=58,tutor-end=59}{1}^{\htmlData{tutor-start=61,tutor-end=62}{2}}} \htmlData{tutor-start=65,tutor-end=66}{=} \htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{1}$。

已知切点坐标和斜率，使用直线的点斜式方程 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y}_{\htmlData{tutor-start=5,tutor-end=6}{0}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{0}}\htmlData{tutor-start=19,tutor-end=20}{)}$，然后整理为一般式以匹配选项。

详细展开： 切点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{0}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}$，斜率 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$。 代入点斜式： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{1}$ 移项整理： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{y} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{1} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{y} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{0}$ 对比选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}$ 结果与 (B) 一致。

根据正四棱柱的几何特征，以底面顶点为原点建立空间直角坐标系，将几何长度转化为坐标数值。

详细展开： 设正四棱柱的底面边长 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{a}$。由题设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}$，可知侧棱长 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{a}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的中点，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{a}$。

以 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 为原点，$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{A}}\htmlData{tutor-start=19,tutor-end=20}{,} \overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{D}\htmlData{tutor-start=38,tutor-end=39}{C}}\htmlData{tutor-start=40,tutor-end=41}{,} \overrightarrow{\htmlData{tutor-start=58,tutor-end=59}{D}\htmlData{tutor-start=59,tutor-end=60}{D}_{\htmlData{tutor-start=62,tutor-end=63}{1}}}$ 的方向分别为 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{z}$ 轴的正方向建立空间直角坐标系 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{z}$。 则各顶点的坐标如下： $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{D}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{)}$

点 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 上，且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{a}$，故 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 坐标与 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 相同，$\htmlData{tutor-start=0,tutor-end=1}{z}$ 坐标为 $\htmlData{tutor-start=0,tutor-end=1}{a}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{)}$。

求出直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的方向向量，利用向量夹角公式计算余弦值，并注意异面直线夹角范围为 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \frac{\htmlData{tutor-start=10,tutor-end=13}{\pi}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{]}$。

详细展开： 1. 计算向量 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}}$： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{E} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{B} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{a}\htmlData{tutor-start=32,tutor-end=33}{,} \htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{,} \htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{)} \htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{a}\htmlData{tutor-start=44,tutor-end=45}{,} \htmlData{tutor-start=46,tutor-end=47}{a}\htmlData{tutor-start=47,tutor-end=48}{,} \htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{)} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{,} \htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{a}\htmlData{tutor-start=60,tutor-end=61}{,} \htmlData{tutor-start=62,tutor-end=63}{a}\htmlData{tutor-start=63,tutor-end=64}{)}$。

2. 计算向量 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{D}_{\htmlData{tutor-start=20,tutor-end=21}{1}}}$： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{D}_{\htmlData{tutor-start=20,tutor-end=21}{1}}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{D}_{\htmlData{tutor-start=29,tutor-end=30}{1}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{C} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{0}\htmlData{tutor-start=43,tutor-end=44}{,} \htmlData{tutor-start=45,tutor-end=46}{2}\htmlData{tutor-start=46,tutor-end=47}{a}\htmlData{tutor-start=47,tutor-end=48}{)} \htmlData{tutor-start=49,tutor-end=50}{-} \htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{0}\htmlData{tutor-start=53,tutor-end=54}{,} \htmlData{tutor-start=55,tutor-end=56}{a}\htmlData{tutor-start=56,tutor-end=57}{,} \htmlData{tutor-start=58,tutor-end=59}{0}\htmlData{tutor-start=59,tutor-end=60}{)} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=64}{(}\htmlData{tutor-start=64,tutor-end=65}{0}\htmlData{tutor-start=65,tutor-end=66}{,} \htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{a}\htmlData{tutor-start=69,tutor-end=70}{,} \htmlData{tutor-start=71,tutor-end=72}{2}\htmlData{tutor-start=72,tutor-end=73}{a}\htmlData{tutor-start=73,tutor-end=74}{)}$。

3. 计算两个向量的数量积： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{C}\htmlData{tutor-start=43,tutor-end=44}{D}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{0} \htmlData{tutor-start=54,tutor-end=61}{\times }\htmlData{tutor-start=61,tutor-end=62}{0} \htmlData{tutor-start=63,tutor-end=64}{+} \htmlData{tutor-start=65,tutor-end=66}{(}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{a}\htmlData{tutor-start=68,tutor-end=69}{)} \htmlData{tutor-start=70,tutor-end=77}{\times }\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=80}{a}\htmlData{tutor-start=80,tutor-end=81}{)} \htmlData{tutor-start=82,tutor-end=83}{+} \htmlData{tutor-start=84,tutor-end=85}{a} \htmlData{tutor-start=86,tutor-end=93}{\times }\htmlData{tutor-start=93,tutor-end=94}{2}\htmlData{tutor-start=94,tutor-end=95}{a} \htmlData{tutor-start=96,tutor-end=97}{=} \htmlData{tutor-start=98,tutor-end=99}{0} \htmlData{tutor-start=100,tutor-end=101}{+} \htmlData{tutor-start=102,tutor-end=103}{a}^{\htmlData{tutor-start=105,tutor-end=106}{2}} \htmlData{tutor-start=108,tutor-end=109}{+} \htmlData{tutor-start=110,tutor-end=111}{2}\htmlData{tutor-start=111,tutor-end=112}{a}^{\htmlData{tutor-start=114,tutor-end=115}{2}} \htmlData{tutor-start=117,tutor-end=118}{=} \htmlData{tutor-start=119,tutor-end=120}{3}\htmlData{tutor-start=120,tutor-end=121}{a}^{\htmlData{tutor-start=123,tutor-end=124}{2}}$。

4. 计算两个向量的模： $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{E}}\htmlData{tutor-start=20,tutor-end=21}{|} \htmlData{tutor-start=22,tutor-end=23}{=} \sqrt{\htmlData{tutor-start=30,tutor-end=31}{0}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{+} \htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{a}\htmlData{tutor-start=41,tutor-end=42}{)}^{\htmlData{tutor-start=44,tutor-end=45}{2}} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{a}^{\htmlData{tutor-start=52,tutor-end=53}{2}}} \htmlData{tutor-start=56,tutor-end=57}{=} \sqrt{\htmlData{tutor-start=64,tutor-end=65}{2}\htmlData{tutor-start=65,tutor-end=66}{a}^{\htmlData{tutor-start=68,tutor-end=69}{2}}} \htmlData{tutor-start=72,tutor-end=73}{=} \sqrt{\htmlData{tutor-start=80,tutor-end=81}{2}}\htmlData{tutor-start=82,tutor-end=83}{a}$ （因 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$）。 $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{D}_{\htmlData{tutor-start=21,tutor-end=22}{1}}}\htmlData{tutor-start=24,tutor-end=25}{|} \htmlData{tutor-start=26,tutor-end=27}{=} \sqrt{\htmlData{tutor-start=34,tutor-end=35}{0}^{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{+} \htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{a}\htmlData{tutor-start=45,tutor-end=46}{)}^{\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=52}{+} \htmlData{tutor-start=53,tutor-end=54}{(}\htmlData{tutor-start=54,tutor-end=55}{2}\htmlData{tutor-start=55,tutor-end=56}{a}\htmlData{tutor-start=56,tutor-end=57}{)}^{\htmlData{tutor-start=59,tutor-end=60}{2}}} \htmlData{tutor-start=63,tutor-end=64}{=} \sqrt{\htmlData{tutor-start=71,tutor-end=72}{a}^{\htmlData{tutor-start=74,tutor-end=75}{2}} \htmlData{tutor-start=77,tutor-end=78}{+} \htmlData{tutor-start=79,tutor-end=80}{4}\htmlData{tutor-start=80,tutor-end=81}{a}^{\htmlData{tutor-start=83,tutor-end=84}{2}}} \htmlData{tutor-start=87,tutor-end=88}{=} \sqrt{\htmlData{tutor-start=95,tutor-end=96}{5}\htmlData{tutor-start=96,tutor-end=97}{a}^{\htmlData{tutor-start=99,tutor-end=100}{2}}} \htmlData{tutor-start=103,tutor-end=104}{=} \sqrt{\htmlData{tutor-start=111,tutor-end=112}{5}}\htmlData{tutor-start=113,tutor-end=114}{a}$。

5. 计算夹角余弦值： 设异面直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 所成的角为 $\htmlData{tutor-start=0,tutor-end=6}{\theta}$，则 $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{|}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{B}\htmlData{tutor-start=38,tutor-end=39}{E}} \htmlData{tutor-start=41,tutor-end=47}{\cdot }\overrightarrow{\htmlData{tutor-start=63,tutor-end=64}{C}\htmlData{tutor-start=64,tutor-end=65}{D}_{\htmlData{tutor-start=67,tutor-end=68}{1}}}\htmlData{tutor-start=70,tutor-end=71}{|}}{\htmlData{tutor-start=73,tutor-end=74}{|}\overrightarrow{\htmlData{tutor-start=90,tutor-end=91}{B}\htmlData{tutor-start=91,tutor-end=92}{E}}\htmlData{tutor-start=93,tutor-end=94}{|} \htmlData{tutor-start=95,tutor-end=101}{\cdot }\htmlData{tutor-start=101,tutor-end=102}{|}\overrightarrow{\htmlData{tutor-start=118,tutor-end=119}{C}\htmlData{tutor-start=119,tutor-end=120}{D}_{\htmlData{tutor-start=122,tutor-end=123}{1}}}\htmlData{tutor-start=125,tutor-end=126}{|}} \htmlData{tutor-start=128,tutor-end=129}{=} \frac{\htmlData{tutor-start=136,tutor-end=137}{3}\htmlData{tutor-start=137,tutor-end=138}{a}^{\htmlData{tutor-start=140,tutor-end=141}{2}}}{\sqrt{\htmlData{tutor-start=150,tutor-end=151}{2}}\htmlData{tutor-start=152,tutor-end=153}{a} \htmlData{tutor-start=154,tutor-end=160}{\cdot }\sqrt{\htmlData{tutor-start=166,tutor-end=167}{5}}\htmlData{tutor-start=168,tutor-end=169}{a}} \htmlData{tutor-start=171,tutor-end=172}{=} \frac{\htmlData{tutor-start=179,tutor-end=180}{3}\htmlData{tutor-start=180,tutor-end=181}{a}^{\htmlData{tutor-start=183,tutor-end=184}{2}}}{\sqrt{\htmlData{tutor-start=193,tutor-end=194}{1}\htmlData{tutor-start=194,tutor-end=195}{0}}\htmlData{tutor-start=196,tutor-end=197}{a}^{\htmlData{tutor-start=199,tutor-end=200}{2}}} \htmlData{tutor-start=203,tutor-end=204}{=} \frac{\htmlData{tutor-start=211,tutor-end=212}{3}}{\sqrt{\htmlData{tutor-start=220,tutor-end=221}{1}\htmlData{tutor-start=221,tutor-end=222}{0}}} \htmlData{tutor-start=225,tutor-end=226}{=} \frac{\htmlData{tutor-start=233,tutor-end=234}{3}\sqrt{\htmlData{tutor-start=240,tutor-end=241}{1}\htmlData{tutor-start=241,tutor-end=242}{0}}}{\htmlData{tutor-start=245,tutor-end=246}{1}\htmlData{tutor-start=246,tutor-end=247}{0}}$。

对比选项，结果为 (C)。

首先根据坐标计算 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 的模，然后对已知等式 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{5}\sqrt{\htmlData{tutor-start=39,tutor-end=40}{2}}$ 两边平方，展开得到关于 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}$ 的方程。

详细展开： 1. 计算 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}$： 已知 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}$，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{2}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{+} \htmlData{tutor-start=33,tutor-end=34}{1}^{\htmlData{tutor-start=36,tutor-end=37}{2}}} \htmlData{tutor-start=40,tutor-end=41}{=} \sqrt{\htmlData{tutor-start=48,tutor-end=49}{5}}$， 故 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{5}$。

2. 展开 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}}$： 由向量运算性质 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{u}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{v}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{u}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{+} \htmlData{tutor-start=61,tutor-end=62}{2}\boldsymbol{\htmlData{tutor-start=74,tutor-end=75}{u}}\htmlData{tutor-start=76,tutor-end=81}{\cdot}\boldsymbol{\htmlData{tutor-start=93,tutor-end=94}{v}} \htmlData{tutor-start=96,tutor-end=97}{+} \htmlData{tutor-start=98,tutor-end=99}{|}\boldsymbol{\htmlData{tutor-start=111,tutor-end=112}{v}}\htmlData{tutor-start=113,tutor-end=114}{|}^{\htmlData{tutor-start=116,tutor-end=117}{2}}$，可得： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{+} \htmlData{tutor-start=61,tutor-end=62}{2}\boldsymbol{\htmlData{tutor-start=74,tutor-end=75}{a}}\htmlData{tutor-start=76,tutor-end=81}{\cdot}\boldsymbol{\htmlData{tutor-start=93,tutor-end=94}{b}} \htmlData{tutor-start=96,tutor-end=97}{+} \htmlData{tutor-start=98,tutor-end=99}{|}\boldsymbol{\htmlData{tutor-start=111,tutor-end=112}{b}}\htmlData{tutor-start=113,tutor-end=114}{|}^{\htmlData{tutor-start=116,tutor-end=117}{2}}$。

3. 代入已知数值： 已知 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{5}\sqrt{\htmlData{tutor-start=41,tutor-end=42}{2}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{5}\sqrt{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{)}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{5} \htmlData{tutor-start=59,tutor-end=66}{\times }\htmlData{tutor-start=66,tutor-end=67}{2} \htmlData{tutor-start=68,tutor-end=69}{=} \htmlData{tutor-start=70,tutor-end=71}{5}\htmlData{tutor-start=71,tutor-end=72}{0}$。 已知 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=19}{\cdot}\boldsymbol{\htmlData{tutor-start=31,tutor-end=32}{b}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{0}$。 已知 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{5}$。

将这些值代入展开式： $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{5} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2} \htmlData{tutor-start=11,tutor-end=18}{\times }\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{|}\boldsymbol{\htmlData{tutor-start=36,tutor-end=37}{b}}\htmlData{tutor-start=38,tutor-end=39}{|}^{\htmlData{tutor-start=41,tutor-end=42}{2}}$。

求解上一步建立的关于 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}}$ 的一元一次方程，并开方得到最终答案。

详细展开： 由方程： $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{5} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{0} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{|}\boldsymbol{\htmlData{tutor-start=27,tutor-end=28}{b}}\htmlData{tutor-start=29,tutor-end=30}{|}^{\htmlData{tutor-start=32,tutor-end=33}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{5} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{|}\boldsymbol{\htmlData{tutor-start=23,tutor-end=24}{b}}\htmlData{tutor-start=25,tutor-end=26}{|}^{\htmlData{tutor-start=28,tutor-end=29}{2}}$ 移项得： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=25}{0} \htmlData{tutor-start=26,tutor-end=27}{-} \htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{5} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{5}$

因为向量的模是非负实数，所以： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{5}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{5}$。

对比选项，结果为 (C)。

分别用底数幂和常见数值夹住三个对数。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{9}$，故 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{2}$；$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{2}$，故 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$；$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{<}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}}$，故 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}$。

把两者改写为自然对数比。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\ln\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{>}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{2}$，而 $\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\ln\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{2}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{c}$。

向右平移 $\htmlData{tutor-start=0,tutor-end=3}{\pi}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{6}$ 要把 x 换成 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{6}$。

详细展开：新相位为 $\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=15}{\omega}\htmlData{tutor-start=15,tutor-end=18}{\pi}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{6}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=24}{\pi}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{4}$，它与 $\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=12}{\pi}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{6}$ 相差整数倍 $\htmlData{tutor-start=0,tutor-end=3}{\pi}$。

解得 $\htmlData{tutor-start=0,tutor-end=6}{\omega}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{k}$。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ 给 $\htmlData{tutor-start=0,tutor-end=6}{\omega}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}$；$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}$ 非正，$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$ 更大。

抛物线 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{x}$ 上点可写成 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{t}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{)}$。

详细展开：代入 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}$ 得 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{t}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}$，两交点参数满足 $\htmlData{tutor-start=0,tutor-end=1}{t}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{t}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}$。焦半径为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{t}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{)}$。

令较短端参数平方为 u。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{u}\htmlData{tutor-start=11,tutor-end=12}{)}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。故两正根为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}$，根和 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{/}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}$。

甲、乙各从四门中选两门。

详细展开：每人有 $\binom42=6$ 种，二人的有序选择共有 $\htmlData{tutor-start=0,tutor-end=1}{6}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{6}$ 种。

不满足“至少一门不同”的仅有两人课程集合完全相同。

详细展开：共同集合有 6 种，所以所求为 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{0}$。

令 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{e}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{e}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}$，并以 F 为分点。

详细展开：设 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{t}\htmlData{tutor-start=6,tutor-end=7}{,}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{t}\htmlData{tutor-start=16,tutor-end=17}{)}$，则由 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{F}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{4}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{F}\htmlData{tutor-start=38,tutor-end=39}{B}}$ 有 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{t}\htmlData{tutor-start=7,tutor-end=8}{,}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{t}\htmlData{tutor-start=17,tutor-end=18}{)}$。把 A、B 分别代入双曲线。

两点方程消去 t 后得到离心率候选。

详细展开：消元因式为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{e}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{e}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}$。$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 对应的交点次序与给定有向比例不符，代回正参数条件留下 $\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{5}$。

以标“上”的面为基准，右邻面是东，左邻面依次是西、下。

详细展开：展开图横向四格绕同一组平行棱折叠，方位依次循环，因此最左横格折成“下”面。

三角形所在格是“下”面在展开图中的上邻格。

详细展开：保持已知上、东的朝向折叠，该格的外法向指北，故三角形面为北面。

选取第二项 k 次。

详细展开：一般项的幂次为 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{k}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{k}}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{k}\htmlData{tutor-start=31,tutor-end=32}{/}\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{y}^{\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{k}\htmlData{tutor-start=40,tutor-end=41}{/}\htmlData{tutor-start=41,tutor-end=42}{2}}$。

令两个指数都等于 3。

详细展开：得到 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$，符号为正，系数是 $\binom42=6$。

设首项为 a1、公差为 d。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{3}}$ 给出 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{d}\htmlData{tutor-start=19,tutor-end=20}{)}$，即 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{d}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}$。

把关系代入等差数列求和公式。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{9}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{8}\htmlData{tutor-start=16,tutor-end=17}{d}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{5}\htmlData{tutor-start=23,tutor-end=24}{d}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{2}$，$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{d}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{5}\htmlData{tutor-start=22,tutor-end=23}{d}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{2}$，比值为 9。

设球半径为 R，OM=R/2。

详细展开：平面与 OA 成 $45^\circ$，故其法线与 OA 也成 $45^\circ$，球心到平面距离 $d=(R/2)\cos45^\circ=R/(2\sqrt{2})$。

截面圆半径平方为 $\htmlData{tutor-start=0,tutor-end=1}{R}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{d}^{\htmlData{tutor-start=9,tutor-end=10}{2}}$。

详细展开：$\htmlData{tutor-start=0,tutor-end=3}{\pi}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{R}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{R}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{7}\htmlData{tutor-start=20,tutor-end=23}{\pi}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{4}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{R}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}$，球表面积为 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=5}{\pi }\htmlData{tutor-start=5,tutor-end=6}{R}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{8}\htmlData{tutor-start=12,tutor-end=15}{\pi}$。

取两弦方向为正交单位向量，M 的分量为 α、β。

详细展开：$\htmlData{tutor-start=0,tutor-end=6}{\alpha}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=16}{\beta}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{O}\htmlData{tutor-start=22,tutor-end=23}{M}^{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{3}$。两条半弦长平方分别为 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=7}{\beta}^{\htmlData{tutor-start=9,tutor-end=10}{2}}$、$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\alpha}^{\htmlData{tutor-start=10,tutor-end=11}{2}}$。

四边形的两条对角线就是两弦且互相垂直。

详细展开：面积 $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{l}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{l}_{\htmlData{tutor-start=11,tutor-end=12}{2}}$，而 $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{l}_{\htmlData{tutor-start=12,tutor-end=13}{2}}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=27}{\alpha}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=36}{\beta}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=43}{\le}\htmlData{tutor-start=43,tutor-end=44}{4}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{9}\htmlData{tutor-start=46,tutor-end=47}{/}\htmlData{tutor-start=47,tutor-end=48}{4}\htmlData{tutor-start=48,tutor-end=49}{=}\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{5}\htmlData{tutor-start=51,tutor-end=52}{/}\htmlData{tutor-start=52,tutor-end=53}{4}$，等号可取。

利用 A+B+C=π 化积。

详细展开：$\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{+}\cos \htmlData{tutor-start=15,tutor-end=16}{B}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}\sin \htmlData{tutor-start=23,tutor-end=24}{A}\sin \htmlData{tutor-start=29,tutor-end=30}{C}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{2}$，所以 $\sin \htmlData{tutor-start=5,tutor-end=6}{A}\sin \htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{4}$。

由正弦定理，$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{c}$ 给出 $\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{=}\sin \htmlData{tutor-start=15,tutor-end=16}{A}\sin \htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{4}$。

详细展开：故 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{3}$；若为后者，则 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{3}$，有 $\sin \htmlData{tutor-start=5,tutor-end=6}{A}\sin \htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=15}{\le}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{4}$，矛盾。

令互相垂直的底边沿两坐标轴，高沿第三轴。

详细展开：取 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{p}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{q}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{A}_{\htmlData{tutor-start=33,tutor-end=34}{1}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{,}\htmlData{tutor-start=41,tutor-end=42}{h}\htmlData{tutor-start=42,tutor-end=43}{)}$，则 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{p}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)}$。

DE 垂直平面 BCC1，因而垂直其中的 BC。

详细展开：$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{p}\htmlData{tutor-start=23,tutor-end=24}{,}\htmlData{tutor-start=24,tutor-end=25}{q}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{)}$，点积为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{p}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{q}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{0}$。边长为正，故 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}$。

设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}$，棱柱高为 h。

详细展开：取 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{D}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{h}\htmlData{tutor-start=28,tutor-end=29}{/}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{)}$。平面 ABD 的法向量可取 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，平面 CBD 的法向量可取 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{h}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}$。

上式给 $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}$，再用平面 BCD 的法向量。

详细展开：$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{C}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{-}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{)}$，与法向量 $\htmlData{tutor-start=0,tutor-end=1}{(}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}$ 的投影比绝对值为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}$，故线面角为 $30^\circ$。

把 n 与 n-1 的条件相减。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}$、$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}$，故 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}}$。

按定义整理。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{n}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{a}_{\htmlData{tutor-start=26,tutor-end=27}{n}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{n}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{b}_{\htmlData{tutor-start=43,tutor-end=44}{n}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{1}}$。又 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}$ 得 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}$，所以 bn 是首项 3、公比 2 的等比数列。

由 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=21}{\cdot}\htmlData{tutor-start=21,tutor-end=22}{2}^{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}}$。

详细展开：令 $\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}^{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}$，则 $\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}$，且 $\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}$。

把等差通项乘回缩放因子。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}^{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{2}$。

甲、乙人数比为 10:5=2:1，总样本量为 3。

详细展开：按分层比例分配为 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}$。

样本量必须为整数且总和为 3。

详细展开：所以甲组抽 2 人，乙组抽 1 人。

甲组 4 女、6 男，不放回抽 2 人。

详细展开：恰一女一男有 $\binom41\binom61=24$ 种，总组合数为 $\binom{10}2=45$。

组合均等可能。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{5}$。