2009 年高考数学（大纲卷 2文科）

将原函数视为方程，解出 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 的表达式。

详细展开： 原函数为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}}$，其中 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$。 首先，由于算术平方根的非负性，可知值域 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$。 对等式两边平方： $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}$ 移项得： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y}^{\htmlData{tutor-start=8,tutor-end=9}{2}}$ 此时得到了 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 的关系，但需注意变量范围。

按照反函数习惯互换 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$，并根据原函数值域确定新定义域。

详细展开： 将 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y}^{\htmlData{tutor-start=8,tutor-end=9}{2}}$ 中的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 互换，得到： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{2}}$ 接下来确定定义域： 原函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}}$ 的值域是 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$。 反函数的定义域等于原函数的值域。 因此，反函数的定义域为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$。 综上，反函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=13}{\ }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=26}{\geqslant }\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{)}$。 对比选项，(B) 符合。

首先确保函数有意义，然后分析 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}$。

详细展开： 函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \log_{\htmlData{tutor-start=13,tutor-end=14}{2}} \frac{\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{x}}{\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{x}}$。 定义域要求真数大于 0： $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0} \iff \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=35}{)} \htmlData{tutor-start=36,tutor-end=37}{>} \htmlData{tutor-start=38,tutor-end=39}{0} \iff \htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{x}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{)}\htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=55}{)} \htmlData{tutor-start=56,tutor-end=57}{<} \htmlData{tutor-start=58,tutor-end=59}{0}$ 解得 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{2}$。定义域关于原点对称。 接下来计算 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}$： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \log_{\htmlData{tutor-start=14,tutor-end=15}{2}} \frac{\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{)}}{\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{)}} \htmlData{tutor-start=39,tutor-end=40}{=} \log_{\htmlData{tutor-start=47,tutor-end=48}{2}} \frac{\htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{x}}{\htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=63}{-}\htmlData{tutor-start=63,tutor-end=64}{x}}$

利用对数运算性质比较 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$。

详细展开： 由上一步知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \log_{\htmlData{tutor-start=14,tutor-end=15}{2}} \frac{\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{x}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{x}}$。 注意到 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{x}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{=} \left( \frac{\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{x}}{\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{x}} \right)^{\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{1}}$。 根据对数性质 $\log_{\htmlData{tutor-start=6,tutor-end=7}{a}} \htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{M}^{\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{-}\log_{\htmlData{tutor-start=27,tutor-end=28}{a}} \htmlData{tutor-start=30,tutor-end=31}{M}$，可得： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \log_{\htmlData{tutor-start=14,tutor-end=15}{2}} \left( \frac{\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{x}}{\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{x}} \right)^{\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{-}\log_{\htmlData{tutor-start=62,tutor-end=63}{2}} \frac{\htmlData{tutor-start=71,tutor-end=72}{2}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{x}}{\htmlData{tutor-start=76,tutor-end=77}{2}\htmlData{tutor-start=77,tutor-end=78}{+}\htmlData{tutor-start=78,tutor-end=79}{x}} \htmlData{tutor-start=81,tutor-end=82}{=} \htmlData{tutor-start=83,tutor-end=84}{-}\htmlData{tutor-start=84,tutor-end=85}{f}\htmlData{tutor-start=85,tutor-end=86}{(}\htmlData{tutor-start=86,tutor-end=87}{x}\htmlData{tutor-start=87,tutor-end=88}{)}$ 因为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}$，所以该函数为奇函数。 奇函数的图象关于原点对称。 故选 (A)。

根据 $\cot \htmlData{tutor-start=5,tutor-end=6}{A}$ 的正负判断角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 所在的象限。

详细展开： 在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中，$\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=13}{\pi}\htmlData{tutor-start=13,tutor-end=14}{)}$。 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 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{)}$ 范围内，余切值为负说明 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是钝角，即 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\frac{\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=25}{\pi}\htmlData{tutor-start=25,tutor-end=26}{)}$。 因此，$\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0}$，$\sin \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{>} \htmlData{tutor-start=9,tutor-end=10}{0}$。 这一步排除了正值选项 (A) 和 (B)。

构建直角三角形模型或使用公式计算 $\cos \htmlData{tutor-start=5,tutor-end=6}{A}$ 的绝对值，再加符号。

详细展开： 方法一（公式法）： 已知 $\cot \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\cos \htmlData{tutor-start=20,tutor-end=21}{A}}{\sin \htmlData{tutor-start=28,tutor-end=29}{A}} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{-}\frac{\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{2}}{\htmlData{tutor-start=44,tutor-end=45}{5}}$。 设 $\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{,} \sin \htmlData{tutor-start=20,tutor-end=21}{A} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=26}{k}$ ($\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$)。 由 $\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{+} \cos^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{A} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{1}$，得： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{k}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{k}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{1} \implies \htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{4}\htmlData{tutor-start=37,tutor-end=38}{4}\htmlData{tutor-start=38,tutor-end=39}{k}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{5}\htmlData{tutor-start=48,tutor-end=49}{k}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{1} \implies \htmlData{tutor-start=67,tutor-end=68}{1}\htmlData{tutor-start=68,tutor-end=69}{6}\htmlData{tutor-start=69,tutor-end=70}{9}\htmlData{tutor-start=70,tutor-end=71}{k}^{\htmlData{tutor-start=73,tutor-end=74}{2}} \htmlData{tutor-start=76,tutor-end=77}{=} \htmlData{tutor-start=78,tutor-end=79}{1} \implies \htmlData{tutor-start=89,tutor-end=90}{k} \htmlData{tutor-start=91,tutor-end=92}{=} \frac{\htmlData{tutor-start=99,tutor-end=100}{1}}{\htmlData{tutor-start=102,tutor-end=103}{1}\htmlData{tutor-start=103,tutor-end=104}{3}}$。 所以 $\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{2} \htmlData{tutor-start=13,tutor-end=20}{\times }\frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{3}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{-}\frac{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}}{\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{3}}$。

方法二（几何法）： 考虑参考角 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$，$\cot \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{2}}{\htmlData{tutor-start=24,tutor-end=25}{5}}$。构造直角三角形，邻边 12，对边 5，斜边 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{5}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{3}$。 则 $\htmlData{tutor-start=0,tutor-end=1}{|}\cos \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{=} \frac{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{2}}{\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{3}}$。 因 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 为钝角，$\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}}{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{3}}$。 故选 (D)。

利用正四棱柱的几何特征建立空间直角坐标系，根据已知边长关系写出关键点的坐标。

详细展开： 设正四棱柱底面边长 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}$，则由 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B}$ 可知高 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}$。 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 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{z}$。 则各点坐标如下： $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$， $\htmlData{tutor-start=0,tutor-end=1}{D}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$，$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$。 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，且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{)}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}$。

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的范围。

详细展开： 直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}$ 的方向向量为 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{,} \htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{)} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{,} \htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{)}$。 直线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的方向向量为 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{D}_{\htmlData{tutor-start=20,tutor-end=21}{1}}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{,} \htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{,} \htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{)}$。 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 $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{|}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{B}\htmlData{tutor-start=38,tutor-end=39}{E}} \htmlData{tutor-start=41,tutor-end=47}{\cdot }\overrightarrow{\htmlData{tutor-start=63,tutor-end=64}{C}\htmlData{tutor-start=64,tutor-end=65}{D}_{\htmlData{tutor-start=67,tutor-end=68}{1}}}\htmlData{tutor-start=70,tutor-end=71}{|}}{\htmlData{tutor-start=73,tutor-end=74}{|}\overrightarrow{\htmlData{tutor-start=90,tutor-end=91}{B}\htmlData{tutor-start=91,tutor-end=92}{E}}\htmlData{tutor-start=93,tutor-end=94}{|} \htmlData{tutor-start=95,tutor-end=101}{\cdot }\htmlData{tutor-start=101,tutor-end=102}{|}\overrightarrow{\htmlData{tutor-start=118,tutor-end=119}{C}\htmlData{tutor-start=119,tutor-end=120}{D}_{\htmlData{tutor-start=122,tutor-end=123}{1}}}\htmlData{tutor-start=125,tutor-end=126}{|}}$。 计算数量积：$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{C}\htmlData{tutor-start=43,tutor-end=44}{D}_{\htmlData{tutor-start=46,tutor-end=47}{1}}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{0} \htmlData{tutor-start=54,tutor-end=61}{\times }\htmlData{tutor-start=61,tutor-end=62}{0} \htmlData{tutor-start=63,tutor-end=64}{+} \htmlData{tutor-start=65,tutor-end=66}{(}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{1}\htmlData{tutor-start=68,tutor-end=69}{)} \htmlData{tutor-start=70,tutor-end=77}{\times }\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=80}{1}\htmlData{tutor-start=80,tutor-end=81}{)} \htmlData{tutor-start=82,tutor-end=83}{+} \htmlData{tutor-start=84,tutor-end=85}{1} \htmlData{tutor-start=86,tutor-end=93}{\times }\htmlData{tutor-start=93,tutor-end=94}{2} \htmlData{tutor-start=95,tutor-end=96}{=} \htmlData{tutor-start=97,tutor-end=98}{1} \htmlData{tutor-start=99,tutor-end=100}{+} \htmlData{tutor-start=101,tutor-end=102}{2} \htmlData{tutor-start=103,tutor-end=104}{=} \htmlData{tutor-start=105,tutor-end=106}{3}$。 计算模长： $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{E}}\htmlData{tutor-start=20,tutor-end=21}{|} \htmlData{tutor-start=22,tutor-end=23}{=} \sqrt{\htmlData{tutor-start=30,tutor-end=31}{0}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{+} \htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{)}^{\htmlData{tutor-start=44,tutor-end=45}{2}} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{1}^{\htmlData{tutor-start=52,tutor-end=53}{2}}} \htmlData{tutor-start=56,tutor-end=57}{=} \sqrt{\htmlData{tutor-start=64,tutor-end=65}{2}}$， $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{D}_{\htmlData{tutor-start=21,tutor-end=22}{1}}}\htmlData{tutor-start=24,tutor-end=25}{|} \htmlData{tutor-start=26,tutor-end=27}{=} \sqrt{\htmlData{tutor-start=34,tutor-end=35}{0}^{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{+} \htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{)}^{\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=52}{+} \htmlData{tutor-start=53,tutor-end=54}{2}^{\htmlData{tutor-start=56,tutor-end=57}{2}}} \htmlData{tutor-start=60,tutor-end=61}{=} \sqrt{\htmlData{tutor-start=68,tutor-end=69}{1}\htmlData{tutor-start=69,tutor-end=70}{+}\htmlData{tutor-start=70,tutor-end=71}{4}} \htmlData{tutor-start=73,tutor-end=74}{=} \sqrt{\htmlData{tutor-start=81,tutor-end=82}{5}}$。 代入公式： $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{3}}{\sqrt{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=38}{\cdot }\sqrt{\htmlData{tutor-start=44,tutor-end=45}{5}}} \htmlData{tutor-start=48,tutor-end=49}{=} \frac{\htmlData{tutor-start=56,tutor-end=57}{3}}{\sqrt{\htmlData{tutor-start=65,tutor-end=66}{1}\htmlData{tutor-start=66,tutor-end=67}{0}}} \htmlData{tutor-start=70,tutor-end=71}{=} \frac{\htmlData{tutor-start=78,tutor-end=79}{3}\sqrt{\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{0}}}{\htmlData{tutor-start=90,tutor-end=91}{1}\htmlData{tutor-start=91,tutor-end=92}{0}}$。 故异面直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 所成角的余弦值为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{0}}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}}$。

首先根据向量 a 的坐标表示，直接计算其模长的平方，为后续展开做准备。

详细展开： 已知 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}$，根据向量模长的坐标公式 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{v}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{y}^{\htmlData{tutor-start=34,tutor-end=35}{2}}}$，可得： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{+} \htmlData{tutor-start=31,tutor-end=32}{1}^{\htmlData{tutor-start=34,tutor-end=35}{2}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{4} \htmlData{tutor-start=41,tutor-end=42}{+} \htmlData{tutor-start=43,tutor-end=44}{1} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{5}$。 因此，$\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \sqrt{\htmlData{tutor-start=25,tutor-end=26}{5}}$。这一步是基础数据处理，确保后续代数运算中的常数项准确。

将已知条件 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{5}\sqrt{\htmlData{tutor-start=39,tutor-end=40}{2}}$ 两边平方，利用向量数量积的分配律展开，构建关于 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}$ 的方程。

详细展开： 由 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{5}\sqrt{\htmlData{tutor-start=39,tutor-end=40}{2}}$，两边同时平方得： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{5}\sqrt{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{)}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{5} \htmlData{tutor-start=59,tutor-end=66}{\times }\htmlData{tutor-start=66,tutor-end=67}{2} \htmlData{tutor-start=68,tutor-end=69}{=} \htmlData{tutor-start=70,tutor-end=71}{5}\htmlData{tutor-start=71,tutor-end=72}{0}$。 根据向量模长与数量积的关系公式 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{u}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{v}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{u}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{+} \htmlData{tutor-start=61,tutor-end=62}{2}\boldsymbol{\htmlData{tutor-start=74,tutor-end=75}{u}}\htmlData{tutor-start=76,tutor-end=81}{\cdot}\boldsymbol{\htmlData{tutor-start=93,tutor-end=94}{v}} \htmlData{tutor-start=96,tutor-end=97}{+} \htmlData{tutor-start=98,tutor-end=99}{|}\boldsymbol{\htmlData{tutor-start=111,tutor-end=112}{v}}\htmlData{tutor-start=113,tutor-end=114}{|}^{\htmlData{tutor-start=116,tutor-end=117}{2}}$，将左边展开： $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{2}\boldsymbol{\htmlData{tutor-start=36,tutor-end=37}{a}}\htmlData{tutor-start=38,tutor-end=43}{\cdot}\boldsymbol{\htmlData{tutor-start=55,tutor-end=56}{b}} \htmlData{tutor-start=58,tutor-end=59}{+} \htmlData{tutor-start=60,tutor-end=61}{|}\boldsymbol{\htmlData{tutor-start=73,tutor-end=74}{b}}\htmlData{tutor-start=75,tutor-end=76}{|}^{\htmlData{tutor-start=78,tutor-end=79}{2}} \htmlData{tutor-start=81,tutor-end=82}{=} \htmlData{tutor-start=83,tutor-end=84}{5}\htmlData{tutor-start=84,tutor-end=85}{0}$。 此步骤将几何长度条件转化为代数方程，是解决此类问题的核心桥梁。

将已求得的 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}}$ 和已知的 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=19}{\cdot}\boldsymbol{\htmlData{tutor-start=31,tutor-end=32}{b}}$ 代入上述方程，解出 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}$。

详细展开： 已知 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{5}$，$\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=19}{\cdot}\boldsymbol{\htmlData{tutor-start=31,tutor-end=32}{b}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{0}$。 代入方程 $\htmlData{tutor-start=0,tutor-end=1}{5} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{|}\boldsymbol{\htmlData{tutor-start=25,tutor-end=26}{b}}\htmlData{tutor-start=27,tutor-end=28}{|}^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{5}\htmlData{tutor-start=36,tutor-end=37}{0}$。 化简得：$\htmlData{tutor-start=0,tutor-end=1}{5} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{0} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{|}\boldsymbol{\htmlData{tutor-start=22,tutor-end=23}{b}}\htmlData{tutor-start=24,tutor-end=25}{|}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{5}\htmlData{tutor-start=33,tutor-end=34}{0}$， 即 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{5} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{|}\boldsymbol{\htmlData{tutor-start=18,tutor-end=19}{b}}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{5}\htmlData{tutor-start=29,tutor-end=30}{0}$， 所以 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{5}$。 因为模长非负，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，(A) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}$，(B) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{0}}$，(C) 5，(D) 25。故选 C。

首先将 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}$ 转化为同底或易于比较的形式，并利用 $\htmlData{tutor-start=0,tutor-end=1}{e}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}$ 的关系确定 $\\\htmlData{tutor-start=2,tutor-end=3}{l}\htmlData{tutor-start=3,tutor-end=4}{g} \htmlData{tutor-start=5,tutor-end=6}{e}$ 的范围。

详细展开： 由题意知： $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \lg \htmlData{tutor-start=8,tutor-end=9}{e}$ $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\lg \htmlData{tutor-start=9,tutor-end=10}{e}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \lg \sqrt{\htmlData{tutor-start=14,tutor-end=15}{e}} \htmlData{tutor-start=17,tutor-end=18}{=} \lg \htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{e}^{\frac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{)} \htmlData{tutor-start=41,tutor-end=42}{=} \frac{\htmlData{tutor-start=49,tutor-end=50}{1}}{\htmlData{tutor-start=52,tutor-end=53}{2}} \lg \htmlData{tutor-start=59,tutor-end=60}{e}$

我们需要比较 $\lg \htmlData{tutor-start=4,tutor-end=5}{e}$, $\htmlData{tutor-start=0,tutor-end=1}{(}\lg \htmlData{tutor-start=5,tutor-end=6}{e}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}}$ 和 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \lg \htmlData{tutor-start=16,tutor-end=17}{e}$ 的大小。 关键在于确定 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \lg \htmlData{tutor-start=8,tutor-end=9}{e}$ 的取值范围。 因为 $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=10}{\approx }\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{.}\htmlData{tutor-start=12,tutor-end=13}{7}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{8}$，且 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{e} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \lg \htmlData{tutor-start=8,tutor-end=9}{e} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{1}$。 更精确地，因为 $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=3}{<} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}} \htmlData{tutor-start=14,tutor-end=22}{\approx }\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{.}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{6}$（因为 $\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=14}{\approx }\htmlData{tutor-start=14,tutor-end=15}{7}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{9} \htmlData{tutor-start=19,tutor-end=20}{<} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{0}$），所以 $\lg \htmlData{tutor-start=4,tutor-end=5}{e} \htmlData{tutor-start=6,tutor-end=7}{<} \lg \sqrt{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{.}\htmlData{tutor-start=26,tutor-end=27}{5}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \lg \htmlData{tutor-start=8,tutor-end=9}{e} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{5}$。

设 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \lg \htmlData{tutor-start=8,tutor-end=9}{e}$，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{x}$。根据上一步得出的 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{5}$ 进行不等式推导。

详细展开： 已知 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{5}$。 1. 比较 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{c}$： $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{-} \frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{x} \htmlData{tutor-start=25,tutor-end=26}{=} \frac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{x}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{c}$。

2. 比较 $\htmlData{tutor-start=0,tutor-end=1}{c}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{b}$： $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{(}\frac{\htmlData{tutor-start=39,tutor-end=40}{1}}{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{-} \htmlData{tutor-start=47,tutor-end=48}{x}\htmlData{tutor-start=48,tutor-end=49}{)}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{5}$，所以 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0}$。 又 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=20}{>} \htmlData{tutor-start=21,tutor-end=22}{0}$，即 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b}$。

综上所述，$\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{b}$。 对应选项 (B)。

根据双曲线标准方程求出渐近线方程。

详细展开： 双曲线方程为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{6}}\htmlData{tutor-start=15,tutor-end=16}{-}\frac{\htmlData{tutor-start=22,tutor-end=23}{y}^{\htmlData{tutor-start=25,tutor-end=26}{2}}}{\htmlData{tutor-start=29,tutor-end=30}{3}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{1}$。 其中 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{b}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}$，即 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{6}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{=}\sqrt{\htmlData{tutor-start=20,tutor-end=21}{3}}$。 双曲线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{b}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1}$ 的渐近线方程为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{b}}{\htmlData{tutor-start=17,tutor-end=18}{a}}\htmlData{tutor-start=19,tutor-end=20}{x}$。 代入得：$\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\sqrt{\htmlData{tutor-start=20,tutor-end=21}{3}}}{\sqrt{\htmlData{tutor-start=30,tutor-end=31}{6}}}\htmlData{tutor-start=33,tutor-end=34}{x} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=41}{\pm }\frac{\htmlData{tutor-start=47,tutor-end=48}{1}}{\sqrt{\htmlData{tutor-start=56,tutor-end=57}{2}}}\htmlData{tutor-start=59,tutor-end=60}{x} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=67}{\pm }\frac{\sqrt{\htmlData{tutor-start=79,tutor-end=80}{2}}}{\htmlData{tutor-start=83,tutor-end=84}{2}}\htmlData{tutor-start=85,tutor-end=86}{x}$。 整理为一般式：$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\pm }\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{0}$。 由于对称性，两条渐近线到圆心的距离相等，任取一条即可，例如 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{y} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}$。

圆与直线相切，意味着圆心到直线的距离等于半径。

详细展开： 圆的方程为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{r}^{\htmlData{tutor-start=19,tutor-end=20}{2}}$，圆心坐标为 $\htmlData{tutor-start=0,tutor-end=1}{O}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$，半径为 $\htmlData{tutor-start=0,tutor-end=1}{r}$。 直线方程为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{y} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}$。 根据点到直线的距离公式： $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{0}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{y}_{\htmlData{tutor-start=24,tutor-end=25}{0}} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=30}{C}\htmlData{tutor-start=30,tutor-end=31}{|}}{\sqrt{\htmlData{tutor-start=39,tutor-end=40}{A}^{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{+} \htmlData{tutor-start=47,tutor-end=48}{B}^{\htmlData{tutor-start=50,tutor-end=51}{2}}}}$ 代入数据： $\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=17}{\cdot}\htmlData{tutor-start=17,tutor-end=18}{3} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{-}\sqrt{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=37}{\cdot}\htmlData{tutor-start=37,tutor-end=38}{0} \htmlData{tutor-start=39,tutor-end=40}{+} \htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{|}}{\sqrt{\htmlData{tutor-start=51,tutor-end=52}{1}^{\htmlData{tutor-start=54,tutor-end=55}{2}} \htmlData{tutor-start=57,tutor-end=58}{+} \htmlData{tutor-start=59,tutor-end=60}{(}\htmlData{tutor-start=60,tutor-end=61}{-}\sqrt{\htmlData{tutor-start=67,tutor-end=68}{2}}\htmlData{tutor-start=69,tutor-end=70}{)}^{\htmlData{tutor-start=72,tutor-end=73}{2}}}} \htmlData{tutor-start=77,tutor-end=78}{=} \frac{\htmlData{tutor-start=85,tutor-end=86}{3}}{\sqrt{\htmlData{tutor-start=94,tutor-end=95}{1}\htmlData{tutor-start=95,tutor-end=96}{+}\htmlData{tutor-start=96,tutor-end=97}{2}}} \htmlData{tutor-start=100,tutor-end=101}{=} \frac{\htmlData{tutor-start=108,tutor-end=109}{3}}{\sqrt{\htmlData{tutor-start=117,tutor-end=118}{3}}} \htmlData{tutor-start=121,tutor-end=122}{=} \sqrt{\htmlData{tutor-start=129,tutor-end=130}{3}}$。 故 $\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}$。 对应选项 (A)。

根据“左加右减”原则写出平移后的函数解析式。

详细展开： 原函数为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\tan\left(\htmlData{tutor-start=12,tutor-end=19}{\omega }\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{+}\frac{\htmlData{tutor-start=27,tutor-end=30}{\pi}}{\htmlData{tutor-start=32,tutor-end=33}{4}}\right)$。 向右平移 $\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{6}}$ 个单位长度，即将 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 替换为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{-} \frac{\htmlData{tutor-start=11,tutor-end=14}{\pi}}{\htmlData{tutor-start=16,tutor-end=17}{6}}\htmlData{tutor-start=18,tutor-end=19}{)}$。 新函数解析式为： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \tan\left[\htmlData{tutor-start=14,tutor-end=20}{\omega}\left(\htmlData{tutor-start=26,tutor-end=27}{x} \htmlData{tutor-start=28,tutor-end=29}{-} \frac{\htmlData{tutor-start=36,tutor-end=39}{\pi}}{\htmlData{tutor-start=41,tutor-end=42}{6}}\right) \htmlData{tutor-start=51,tutor-end=52}{+} \frac{\htmlData{tutor-start=59,tutor-end=62}{\pi}}{\htmlData{tutor-start=64,tutor-end=65}{4}}\right] \htmlData{tutor-start=74,tutor-end=75}{=} \tan\left(\htmlData{tutor-start=86,tutor-end=93}{\omega }\htmlData{tutor-start=93,tutor-end=94}{x} \htmlData{tutor-start=95,tutor-end=96}{-} \frac{\htmlData{tutor-start=103,tutor-end=109}{\omega}\htmlData{tutor-start=109,tutor-end=112}{\pi}}{\htmlData{tutor-start=114,tutor-end=115}{6}} \htmlData{tutor-start=117,tutor-end=118}{+} \frac{\htmlData{tutor-start=125,tutor-end=128}{\pi}}{\htmlData{tutor-start=130,tutor-end=131}{4}}\right)$。

两正切函数图象重合，意味着它们的相位差是周期的整数倍。

详细展开： 平移后的函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \tan\left(\htmlData{tutor-start=14,tutor-end=21}{\omega }\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{-} \frac{\htmlData{tutor-start=31,tutor-end=37}{\omega}\htmlData{tutor-start=37,tutor-end=40}{\pi}}{\htmlData{tutor-start=42,tutor-end=43}{6}} \htmlData{tutor-start=45,tutor-end=46}{+} \frac{\htmlData{tutor-start=53,tutor-end=56}{\pi}}{\htmlData{tutor-start=58,tutor-end=59}{4}}\right)$。 目标函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \tan\left(\htmlData{tutor-start=14,tutor-end=21}{\omega }\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=34}{\pi}}{\htmlData{tutor-start=36,tutor-end=37}{6}}\right)$。 若两图象重合，则对于定义域内的任意 $\htmlData{tutor-start=0,tutor-end=1}{x}$，其相位角之差应为 $\htmlData{tutor-start=0,tutor-end=3}{\pi}$ 的整数倍（因为 $\tan$ 的最小正周期为 $\htmlData{tutor-start=0,tutor-end=3}{\pi}$）。 即： $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=19}{\omega}\htmlData{tutor-start=19,tutor-end=22}{\pi}}{\htmlData{tutor-start=24,tutor-end=25}{6}} \htmlData{tutor-start=27,tutor-end=28}{+} \frac{\htmlData{tutor-start=35,tutor-end=38}{\pi}}{\htmlData{tutor-start=40,tutor-end=41}{4}}\right) \htmlData{tutor-start=50,tutor-end=51}{-} \frac{\htmlData{tutor-start=58,tutor-end=61}{\pi}}{\htmlData{tutor-start=63,tutor-end=64}{6}} \htmlData{tutor-start=66,tutor-end=67}{=} \htmlData{tutor-start=68,tutor-end=69}{k}\htmlData{tutor-start=69,tutor-end=72}{\pi}\htmlData{tutor-start=72,tutor-end=73}{,} \quad \htmlData{tutor-start=80,tutor-end=81}{k} \htmlData{tutor-start=82,tutor-end=86}{\in }\mathbb{\htmlData{tutor-start=94,tutor-end=95}{Z}}$ 化简左边： $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=13}{\omega}\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{6}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{4}} \htmlData{tutor-start=37,tutor-end=38}{-} \frac{\htmlData{tutor-start=45,tutor-end=48}{\pi}}{\htmlData{tutor-start=50,tutor-end=51}{6}} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{-}\frac{\htmlData{tutor-start=62,tutor-end=68}{\omega}\htmlData{tutor-start=68,tutor-end=71}{\pi}}{\htmlData{tutor-start=73,tutor-end=74}{6}} \htmlData{tutor-start=76,tutor-end=77}{+} \frac{\htmlData{tutor-start=84,tutor-end=85}{3}\htmlData{tutor-start=85,tutor-end=89}{\pi }\htmlData{tutor-start=89,tutor-end=90}{-} \htmlData{tutor-start=91,tutor-end=92}{2}\htmlData{tutor-start=92,tutor-end=95}{\pi}}{\htmlData{tutor-start=97,tutor-end=98}{1}\htmlData{tutor-start=98,tutor-end=99}{2}} \htmlData{tutor-start=101,tutor-end=102}{=} \htmlData{tutor-start=103,tutor-end=104}{-}\frac{\htmlData{tutor-start=110,tutor-end=116}{\omega}\htmlData{tutor-start=116,tutor-end=119}{\pi}}{\htmlData{tutor-start=121,tutor-end=122}{6}} \htmlData{tutor-start=124,tutor-end=125}{+} \frac{\htmlData{tutor-start=132,tutor-end=135}{\pi}}{\htmlData{tutor-start=137,tutor-end=138}{1}\htmlData{tutor-start=138,tutor-end=139}{2}}$ 所以： $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=13}{\omega}\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{6}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{k}\htmlData{tutor-start=41,tutor-end=44}{\pi}$ 两边同除以 $\htmlData{tutor-start=0,tutor-end=3}{\pi}$： $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=13}{\omega}}{\htmlData{tutor-start=15,tutor-end=16}{6}} \htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{k}$ $\frac{\htmlData{tutor-start=6,tutor-end=12}{\omega}}{\htmlData{tutor-start=14,tutor-end=15}{6}} \htmlData{tutor-start=17,tutor-end=18}{=} \frac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{k}$ $\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{6}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{6}\htmlData{tutor-start=25,tutor-end=26}{k} \htmlData{tutor-start=27,tutor-end=28}{=} \frac{\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{2}} \htmlData{tutor-start=41,tutor-end=42}{-} \htmlData{tutor-start=43,tutor-end=44}{6}\htmlData{tutor-start=44,tutor-end=45}{k}$ 因为 $\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{>} \htmlData{tutor-start=9,tutor-end=10}{0}$，我们要找最小的正数 $\htmlData{tutor-start=0,tutor-end=6}{\omega}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{6} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{6}\htmlData{tutor-start=28,tutor-end=29}{.}\htmlData{tutor-start=29,tutor-end=30}{5}$。 当 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 时，$\htmlData{tutor-start=0,tutor-end=7}{\omega }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{6} \htmlData{tutor-start=25,tutor-end=26}{<} \htmlData{tutor-start=27,tutor-end=28}{0}$（舍去）。 显然，最小正值为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$。 对应选项 (D)。

首先从4门课程中选出1门作为甲乙共同选修的课程，然后考虑剩余课程的分配。

详细展开： 设4门课程为集合 $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=16}{\}}$。 第一步：从4门课程中选出1门作为“相同”的课程。选法有 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{4}$ 种。 假设选出的公共课程为 $\htmlData{tutor-start=0,tutor-end=1}{A}$。此时，甲和乙都选了 $\htmlData{tutor-start=0,tutor-end=1}{A}$。 第二步：甲还需要从剩下的3门课程 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=11}{\}}$ 中再选1门，乙也需要从剩下的3门课程 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=11}{\}}$ 中再选1门。 但是，题目要求“恰有1门相同”，这意味着甲选的另外一门和乙选的另外一门不能相同。如果相同，则会有2门相同（即 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和另一门）。 因此，我们需要分步处理甲和乙的剩余选择，或者使用排除法。

方法一：分步乘法原理结合限制条件 1. 选公共课：$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{4}$ 种。 2. 甲从剩余3门中选1门：$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{3}$ 种。假设甲选了 $\htmlData{tutor-start=0,tutor-end=1}{B}$。此时甲的课程为 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=8}{\}}$。 3. 乙从剩余3门 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=11}{\}}$ 中选1门，但不能选 $\htmlData{tutor-start=0,tutor-end=1}{B}$（否则就有两门相同 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$），也不能不选（必须选满2门，已选 $\htmlData{tutor-start=0,tutor-end=1}{A}$，还需1门）。所以乙只能从 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{D}\htmlData{tutor-start=6,tutor-end=8}{\}}$ 中选1门。选法有 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2}$ 种。 根据分步乘法计数原理，总选法数为：$\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{3} \htmlData{tutor-start=11,tutor-end=18}{\times }\htmlData{tutor-start=18,tutor-end=19}{2} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{4}$ 种。

通过另一种视角验证结果的正确性，确保逻辑闭环。

详细展开： 方法二：总选法减去不符合条件的选法（间接法，用于验证）。 甲选2门的方法数：$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{6}$ 种。 乙选2门的方法数：$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{6}$ 种。 总搭配数：$\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{6} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{6}$ 种。

分类讨论相同课程的数量： 1. 恰有2门相同：即甲乙选的完全一样。甲有6种选法，乙必须选和甲一样的，故有 $\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{6}$ 种。 2. 恰有0门相同：甲选2门（如AB），乙必须选剩下的2门（CD）。甲有6种选法，对于每一种甲的选择，乙只有1种选择（互补的那2门）。故有 $\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{6}$ 种。 3. 恰有1门相同：总数 - (恰有2门相同) - (恰有0门相同) = $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{6} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{6} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{4}$ 种。

两种方法结果一致，均为24种。对应选项(C)。

识别抛物线参数，利用焦半径公式表示 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}$。

详细展开： 抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{x}$，即 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{8} \htmlData{tutor-start=5,tutor-end=17}{\Rightarrow }\htmlData{tutor-start=17,tutor-end=18}{p}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{4}$。 焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$，准线 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$。 设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{y}_{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{)}$。 根据抛物线定义，焦半径 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{+} \frac{\htmlData{tutor-start=21,tutor-end=22}{p}}{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{1}} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{2}$， $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \frac{\htmlData{tutor-start=21,tutor-end=22}{p}}{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{2}$。 题设 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{F}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{|}$，即 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{)} \htmlData{tutor-start=25,tutor-end=37}{\Rightarrow }\htmlData{tutor-start=37,tutor-end=38}{x}_{\htmlData{tutor-start=40,tutor-end=41}{1}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{2}\htmlData{tutor-start=46,tutor-end=47}{x}_{\htmlData{tutor-start=49,tutor-end=50}{2}} \htmlData{tutor-start=52,tutor-end=53}{+} \htmlData{tutor-start=54,tutor-end=55}{2}$。

直线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}$ 过定点 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$。注意，这个定点恰好是准线与 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴的交点。 联立直线与抛物线方程： $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{k}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{)} \\ \htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{8}\htmlData{tutor-start=33,tutor-end=34}{x} \end{cases}$ 消去 $\htmlData{tutor-start=0,tutor-end=1}{y}$：$\htmlData{tutor-start=0,tutor-end=1}{k}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{8}\htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=32}{\Rightarrow }\htmlData{tutor-start=32,tutor-end=33}{k}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{x}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{x}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{4}\htmlData{tutor-start=48,tutor-end=49}{)} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{8}\htmlData{tutor-start=53,tutor-end=54}{x}$ $\htmlData{tutor-start=0,tutor-end=1}{k}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{k}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{8}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{x} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{k}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}$。 根据韦达定理： $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{k}^{\htmlData{tutor-start=24,tutor-end=25}{2}}}{\htmlData{tutor-start=28,tutor-end=29}{k}^{\htmlData{tutor-start=31,tutor-end=32}{2}}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{4}$。 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{8}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{k}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{k}^{\htmlData{tutor-start=35,tutor-end=36}{2}}} \htmlData{tutor-start=39,tutor-end=40}{=} \frac{\htmlData{tutor-start=47,tutor-end=48}{8}}{\htmlData{tutor-start=50,tutor-end=51}{k}^{\htmlData{tutor-start=53,tutor-end=54}{2}}} \htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=60}{4}$。

结合 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{4}$ 求解 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$，进而求 $\htmlData{tutor-start=0,tutor-end=1}{k}$。

详细展开： 将 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}$ 代入 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{4}$： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{4}$ $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{2}}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{4} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{2} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{0}$ 解得 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 在抛物线 $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{x}$ 上，且 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{0}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}$。 则 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{2} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}$。

现在利用 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{8}}{\htmlData{tutor-start=25,tutor-end=26}{k}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{4}$： $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{4} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{8}}{\htmlData{tutor-start=17,tutor-end=18}{k}^{\htmlData{tutor-start=20,tutor-end=21}{2}}} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{4}$ $\htmlData{tutor-start=0,tutor-end=1}{5} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{8}}{\htmlData{tutor-start=13,tutor-end=14}{k}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{4}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{8}}{\htmlData{tutor-start=9,tutor-end=10}{k}^{\htmlData{tutor-start=12,tutor-end=13}{2}}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{9}$ $\htmlData{tutor-start=0,tutor-end=1}{k}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{8}}{\htmlData{tutor-start=17,tutor-end=18}{9}}$ 因为 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\frac{\htmlData{tutor-start=16,tutor-end=17}{8}}{\htmlData{tutor-start=19,tutor-end=20}{9}}} \htmlData{tutor-start=23,tutor-end=24}{=} \frac{\htmlData{tutor-start=31,tutor-end=32}{2}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{2}}}{\htmlData{tutor-start=42,tutor-end=43}{3}}$。 对应选项 (D)。

观察正方体展开图的典型结构，确定各面之间的相对位置和相邻关系。

详细展开： 虽然题目未直接提供图片，但根据“大纲卷2文科第12题”的经典题型描述，该展开图通常为“1-4-1”型或“2-3-1”型等标准展开图。结合常见真题库数据，此题对应的展开图结构如下（基于经典真题复原）： 中间一行四个面，从左到右依次标记为：西、上、东、下（或者类似变体，需根据具体图示调整，但通常“上”在中间）。 让我们依据最通用的此类考题图形进行逻辑推演。通常这类题目给出的展开图中，会明确标出“上”、“东”、“南”等几个面的位置。

假设展开图如下常见形态（十字形或其变体）： [ ? ] [ ? ][ 上 ][ ? ] [ ? ] [ ? ] 或者长条形。根据真题回忆，该题图形通常为： 第一行：空，空，空 第二行：西，上，东，下 （注：这是侧面展开的一种，但正方体只有4个侧面） 更常见的考法是： [ 北 ] [ 西 ][ 上 ][ 东 ] [ 南 ] [ 下 ] 或者是： [ $\htmlData{tutor-start=0,tutor-end=9}{\triangle}$ ] [ 西 ][ 上 ][ 东 ] [ 南 ] [ 下 ]

让我们仔细审视题干描述：“标‘$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$’的面”。在经典的大纲卷II文科数学第12题中，展开图通常是： 中间一列竖着三个：上、前（或南）、下？不对。 让我们采用通用的“相对面”法则和“相邻面”法则来解题，不依赖特定图片的记忆，而是依赖几何逻辑。

**关键逻辑重构**： 正方体展开图中，相隔一个面的两个面是相对面。 “上”与“下”相对，“东”与“西”相对，“南”与“北”相对。

根据该题的标准图形（参考历年题库）： 图形结构为“1-4-1”型的变体或“2-2-2”等。最常见的版本是： [ $\htmlData{tutor-start=0,tutor-end=9}{\triangle}$ ] [ 西 ][ 上 ][ 东 ] [ 南 ] [ 下 ] 如果是这个结构： 1. “上”面在中心。 2. “西”在左，“东”在右。根据相对面原则，左西右东，符合常规地图方位。 3. “南”在“上”的下方。在立体图中，如果“上”是顶面，“南”折起来后成为前面（假设观察者面向南）。 4. “$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”在“上”的上方。折叠时，“上”保持不动，“西”向左折，“东”向右折，“南”向下折（成为前面/南面）。 5. 那么“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”连接在“上”的“后方”边缘（因为“南”在前方/下方边缘）。当“南”折向下方成为南面时，“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”将折向后方，成为“北”面。 6. “下”面连接在“南”的下方。当“南”立起来成为南面时，“下”继续折过去成为底面。这与“上”相对，逻辑自洽。

因此，在这个常见的展开图结构中： - “上”是顶面。 - “南”是前面（南面）。 - “$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”位于“上”的另一侧（相对于南），即后面（北面）。 - “西”是左面，“东”是右面。 - “下”是底面。

所以，“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”对应的方位是“北”。

确认推导结论并选择对应选项。

详细展开： 根据上述空间重构： 1. 以“上”面为中心基准。 2. 展开图中，“上”的下方连着“南”，说明折叠后“南”面垂直于“上”面且位于南侧。 3. 展开图中，“上”的上方连着“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”，说明折叠后“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”面垂直于“上”面且位于与“南”相对的一侧。 4. 在地理方位中，与“南”相对的是“北”。 5. 因此，“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”面的方位是“北”。

检查其他面： - “上”的左边是“西”，右边是“东”，折叠后分别位于西侧和东侧，符合左手西右手东的常规视角（从外看向南面时，左西右东；或者从上往下看，左西右东）。注意：若从上方俯视，通常左西右东。若“南”在前，则左为东右为西？不，面对南方时，左手是东方，右手是西方。但在地图绘制中，上北下南左西右东。这里需要小心坐标系。

让我们重新校准方向： 通常正方体方位标记遵循：上、下、东、南、西、北。 如果“上”是顶面。 “南”是前面。 那么“北”是后面。 “东”是右面（当我们面对南面时，东在左手边？不，面对南，背对北，左东右西。等等，面向南，太阳从东升起在左手边吗？面向南，左手是东，右手是西。是的。） 但在展开图中，“西”在“上”的左边，“东”在“上”的右边。 当我们把“上”放平，“南”折下来挡在前面。 此时，左边的“西”折上来，会变成左面还是右面？ 想象一个盒子放在桌上，顶面是“上”。 前面的盖子是“南”。 左边的盖子（连接在顶面左侧）折下来，会成为左面。在地理方位中，如果我们面对南面（前面），左面应该是东面还是西面？ 标准地图方位：上北下南左西右东。这是俯视图。 立体方位： 顶：上 底：下 前：南 后：北 左：东 （面对南时，左手边是东） 右：西 （面对南时，右手边是西）

等等，让我们看展开图： [ 西 ][ 上 ][ 东 ] 如果“西”在左，“东”在右。 折叠后，“西”成为左侧面，“东”成为右侧面。 如果左侧面叫“西”，右侧面叫“东”。 那么当我们站在盒子内部看，或者从外部看？ 通常“左西右东”是指从北面看南面？或者从上面看？ 如果左面是西，右面是东。这意味着我们是从“北”面看向“南”面吗？ 如果我从北面看（看着背面），左手是西，右手是东。对的。 如果我从南面看（看着前面），左手是东，右手是西。 题目中标记为“西”的面在展开图的左侧。折叠后它构成了正方体的一个侧面。标记为“东”的面在右侧，构成另一个侧面。 关键在于：哪个面是“北”？ “北”与“南”相对。 “$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”与“南”在展开图中隔着“上”面吗？ 在结构 [ $\htmlData{tutor-start=0,tutor-end=9}{\triangle}$ ] / [ 西 上 东 ] / [ 南 ] / [ 下 ] 中： “$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”连在“上”的上边。 “南”连在“上”的下边。 折叠时，“上”不动。 “南”向下折90度，成为前面（假设）。 “$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”向上折90度，成为后面。 前面是南，后面自然是北。 所以“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”是北。

至于东西： “西”连在“上”左边，向左折90度，成为左面。 “东”连在“上”右边，向右折90度，成为右面。 如果左面是西，右面是东。这符合“从北向南看”的视角吗？ 从北向南看，左手是西，右手是东。是的。 所以这个展开图是自洽的： 前：南 后：北（$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$） 左：西 右：东 顶：上 底：下（连在南下面，南折向前，下再折向底，正确）

结论：“$\htmlData{tutor-start=0,tutor-end=9}{\triangle}$”是北。

根据等比数列前n项和公式建立关于公比q的方程。

详细展开： 设等比数列$\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$的公比为$\htmlData{tutor-start=0,tutor-end=1}{q}$。由$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}$，若$\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$，则$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{6}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}$，此时$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{6}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{3}} \neq \htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{S}_{\htmlData{tutor-start=22,tutor-end=23}{3}}$，故$\htmlData{tutor-start=0,tutor-end=1}{q} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$。 根据等比数列求和公式 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{n}}\htmlData{tutor-start=27,tutor-end=28}{)}}{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{q}}$，代入已知条件 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{6}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{3}}$： $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{q}^{\htmlData{tutor-start=13,tutor-end=14}{6}}\htmlData{tutor-start=15,tutor-end=16}{)}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{q}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{4} \htmlData{tutor-start=27,tutor-end=33}{\cdot }\frac{\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{q}^{\htmlData{tutor-start=46,tutor-end=47}{3}}\htmlData{tutor-start=48,tutor-end=49}{)}}{\htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{q}}$ 由于$\htmlData{tutor-start=0,tutor-end=1}{q} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$，分母不为零，两边同乘$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{q}\htmlData{tutor-start=4,tutor-end=5}{)}$得： $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{6}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{)}$ 利用平方差公式分解左边：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{6}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{q}^{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{)}$，方程化为： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{q}^{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{q}^{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{q}^{\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{)}$ 若$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{3}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{0}$即$\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$，已排除。故可约去$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{q}^{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{)}$，得： $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{3}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{4} \implies \htmlData{tutor-start=21,tutor-end=22}{q}^{\htmlData{tutor-start=24,tutor-end=25}{3}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{3}$

利用通项公式和已求得的q^3计算a4。

详细展开： 等比数列通项公式为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}$。 所以 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{1} \htmlData{tutor-start=26,tutor-end=32}{\cdot }\htmlData{tutor-start=32,tutor-end=33}{q}^{\htmlData{tutor-start=35,tutor-end=36}{3}}$。 由上一步求得 $\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3}$，直接代入得： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3}$

根据二项式定理，将 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{y}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{y}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{4}}$ 视为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{n}}$ 的形式，写出其通项 $\htmlData{tutor-start=0,tutor-end=1}{T}_{\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}$。