2008 年高考数学（北京卷文科）

首先观察 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{.}\htmlData{tutor-start=15,tutor-end=16}{8}$，通过真数与 1 的关系判断其正负。

详细展开： 对于 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{.}\htmlData{tutor-start=15,tutor-end=16}{8}$： 因为底数 $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$，对数函数 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{x}$ 单调递增。 真数 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{8} \htmlData{tutor-start=4,tutor-end=5}{<} \htmlData{tutor-start=6,tutor-end=7}{1}$，所以 $\log_{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{8} \htmlData{tutor-start=13,tutor-end=14}{<} \log_{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{1} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{0}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$。 这是一个非常关键的切入点，通常对数比较大小中，负数是最小的候选者之一。

接下来分析 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=16}{\pi}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{7}} \htmlData{tutor-start=13,tutor-end=14}{6}$，确定它们是否为正数以及是否大于 1。

详细展开： 对于 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=16}{\pi}$： 因为 $\htmlData{tutor-start=0,tutor-end=4}{\pi }\htmlData{tutor-start=4,tutor-end=12}{\approx }\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{4} \htmlData{tutor-start=17,tutor-end=18}{>} \htmlData{tutor-start=19,tutor-end=20}{3}$，且底数 $\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$，所以 $\log_{\htmlData{tutor-start=6,tutor-end=7}{3}} \htmlData{tutor-start=9,tutor-end=13}{\pi }\htmlData{tutor-start=13,tutor-end=14}{>} \log_{\htmlData{tutor-start=21,tutor-end=22}{3}} \htmlData{tutor-start=24,tutor-end=25}{3} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{1}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$。

对于 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{7}} \htmlData{tutor-start=13,tutor-end=14}{6}$： 因为 $\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{7}$，且底数 $\htmlData{tutor-start=0,tutor-end=1}{7} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$，所以 $\log_{\htmlData{tutor-start=6,tutor-end=7}{7}} \htmlData{tutor-start=9,tutor-end=10}{6} \htmlData{tutor-start=11,tutor-end=12}{<} \log_{\htmlData{tutor-start=19,tutor-end=20}{7}} \htmlData{tutor-start=22,tutor-end=23}{7} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{1}$。 同时，因为 $\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$，所以 $\log_{\htmlData{tutor-start=6,tutor-end=7}{7}} \htmlData{tutor-start=9,tutor-end=10}{6} \htmlData{tutor-start=11,tutor-end=12}{>} \log_{\htmlData{tutor-start=19,tutor-end=20}{7}} \htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0}$。 即 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{1}$。

综合以上分析： $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$ $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{1}$ $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$

因此，大小关系必然为 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{c}$。

假设双曲线方程为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{y}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{6}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1}$，计算其准线方程，看是否与给定条件一致。

详细展开： 对于双曲线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{-} \frac{\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{b}^{\htmlData{tutor-start=38,tutor-end=39}{2}}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{1}$，参数为 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{b}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{6}$。 则 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{b} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{4}$。 半焦距 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{b}^{\htmlData{tutor-start=21,tutor-end=22}{2}}} \htmlData{tutor-start=25,tutor-end=26}{=} \sqrt{\htmlData{tutor-start=33,tutor-end=34}{9} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{6}} \htmlData{tutor-start=41,tutor-end=42}{=} \sqrt{\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{5}} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{5}$。 焦点在 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴上，准线方程公式为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}{\htmlData{tutor-start=21,tutor-end=22}{c}}$。 代入数值：$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{9}}{\htmlData{tutor-start=17,tutor-end=18}{5}}$。 这与条件“双曲线的准线方程为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{9}}{\htmlData{tutor-start=17,tutor-end=18}{5}}$”完全一致。 因此，充分性成立。

假设双曲线的准线方程为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{9}}{\htmlData{tutor-start=17,tutor-end=18}{5}}$，判断其方程是否唯一确定为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{y}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{6}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1}$。

详细展开： 准线方程 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\frac{\htmlData{tutor-start=14,tutor-end=15}{9}}{\htmlData{tutor-start=17,tutor-end=18}{5}}$ 暗示焦点在 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴上，设双曲线方程为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{-} \frac{\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{b}^{\htmlData{tutor-start=38,tutor-end=39}{2}}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{1}$ ($\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}$)。 由准线公式知 $\frac{\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{c}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{9}}{\htmlData{tutor-start=27,tutor-end=28}{5}}$，即 $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{c}$。 又 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{b}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{9}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{b}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}$。 这是一个关于 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 的方程，有无数多组解。 例如： 若取 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$，则 $45 = 9\sqrt{9+b^{2}} \implies 5 = \sqrt{9+b^{2}} \implies 25 = 9+b^{2} \implies b^{2}=16$，此时方程为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{-} \frac{\htmlData{tutor-start=24,tutor-end=25}{y}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{6}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1}$。 但若取其他满足 $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{9}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{b}^{\htmlData{tutor-start=25,tutor-end=26}{2}}}$ 的 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}$ 值，比如缩放比例不同的双曲线，只要保持 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{c}$ 比值不变即可？ 等等，让我们更仔细地分析。准线只确定了 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{c}$ 的值。离心率 $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{a}$。$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{e} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{c} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{9}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{5}$。 这意味着 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{9}}{\htmlData{tutor-start=13,tutor-end=14}{5}} \htmlData{tutor-start=16,tutor-end=17}{e}$。而 $\htmlData{tutor-start=0,tutor-end=1}{e} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$。 $\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{c}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{a}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{e}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{)}$。 方程可以写为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}} \htmlData{tutor-start=20,tutor-end=21}{-} \frac{\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{a}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{e}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{)}} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{1}$。 只要 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{e}$ 满足上述关系，准线都是 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{5}$。 显然，$\htmlData{tutor-start=0,tutor-end=1}{a}$ 可以取不同的值（对应不同的 $\htmlData{tutor-start=0,tutor-end=1}{e}$），从而得到不同的双曲线方程。 例如，若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$，则 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{c} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{5} \implies \htmlData{tutor-start=19,tutor-end=20}{c}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{5}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{9}$，但这要求 $\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{a}$，即 $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{9} \htmlData{tutor-start=4,tutor-end=5}{>} \htmlData{tutor-start=6,tutor-end=7}{1}$ 不成立。所以 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 不能任意小。 由 $c > a \implies \frac{5a^{2}}{9} > a \implies a > \frac{9}{5}$。 取 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{9}}{\htmlData{tutor-start=13,tutor-end=14}{5}} \htmlData{tutor-start=16,tutor-end=23}{\times }\sqrt{\htmlData{tutor-start=29,tutor-end=30}{2}}$ (举例)，则 $\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{5}}{\htmlData{tutor-start=13,tutor-end=14}{9}} \htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \frac{\htmlData{tutor-start=30,tutor-end=31}{5}}{\htmlData{tutor-start=33,tutor-end=34}{9}} \htmlData{tutor-start=36,tutor-end=42}{\cdot }\frac{\htmlData{tutor-start=48,tutor-end=49}{8}\htmlData{tutor-start=49,tutor-end=50}{1}}{\htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=54}{5}} \htmlData{tutor-start=56,tutor-end=62}{\cdot }\htmlData{tutor-start=62,tutor-end=63}{2} \htmlData{tutor-start=64,tutor-end=65}{=} \frac{\htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{8}}{\htmlData{tutor-start=76,tutor-end=77}{5}}$? 不对，$\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{/} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{5}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{5}\htmlData{tutor-start=21,tutor-end=22}{a}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{9}$。 总之，存在无数个不同的 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 组合使得准线为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=8}{\pm }\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{5}$。因此，原方程不是唯一的。 所以，必要性不成立。

根据正弦定理 $\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\sin \htmlData{tutor-start=14,tutor-end=15}{A}} \htmlData{tutor-start=17,tutor-end=18}{=} \frac{\htmlData{tutor-start=25,tutor-end=26}{b}}{\sin \htmlData{tutor-start=33,tutor-end=34}{B}}$，代入已知数据求解 $\sin \htmlData{tutor-start=5,tutor-end=6}{A}$。

详细展开： 已知 $a = \sqrt{2}, b = \sqrt{3}, B = 60^\circ$。 由正弦定理： $\frac{\sqrt{2}}{\sin A} = \frac{\sqrt{3}}{\sin 60^\circ}$ $\sin A = \frac{\sqrt{2} \cdot \sin 60^\circ}{\sqrt{3}} = \frac{\sqrt{2} \cdot \frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{\sqrt{2}}{2}$ 因为 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是三角形内角，且 $\sin \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{2}}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 可能为 $45^\circ$ 或 $135^\circ$。

根据三角形中“大边对大角”的性质，判断 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的取值范围，从而排除增根。

详细展开： 已知 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{b} \htmlData{tutor-start=16,tutor-end=17}{=} \sqrt{\htmlData{tutor-start=24,tutor-end=25}{3}}$。 因为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{<} \sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}$，即 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{B}$。 已知 $B = 60^\circ$，所以 $A < 60^\circ$。 在前一步得到的两个候选值 $45^\circ$ 和 $135^\circ$ 中： $135^\circ > 60^\circ$，不符合 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{B}$，舍去。 $45^\circ < 60^\circ$，符合条件。 此外，若 $A=135^\circ$，则 $A+B = 135^\circ+60^\circ = 195^\circ > 180^\circ$，构不成三角形，也可由此排除。 因此，$A = 45^\circ$。

首先分析原函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在给定定义域 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}$ 上的单调性和值域。反函数的定义域等于原函数的值域，这是确定反函数中自变量取值范围的关键依据。

详细展开： 原函数为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{1}$。 这是一个开口向上的抛物线，对称轴为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}$。 题目给定的定义域为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}$，即对称轴的左侧。 在区间 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上，随着 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的增大，$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$ 减小，因此 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 单调递减。 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\to }\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=13}{\infty}$ 时，$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=9}{\to }\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=16}{\infty}$。 当 $x \to 1^-$（从左侧趋近于 1）时，$(x-1)^{2} \to 0^+$，故 $f(x) \to 1^+$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{>} \htmlData{tutor-start=12,tutor-end=13}{0}$，从而 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{1} \htmlData{tutor-start=21,tutor-end=22}{>} \htmlData{tutor-start=23,tutor-end=24}{1}$。 因此，原函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的值域为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$。 根据反函数的性质，反函数 $\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}$ 的定义域即为原函数的值域，故 $\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}$ 的定义域为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$。 据此可以排除选项 (C) 和 (D)，因为它们的定义域包含 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$（即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{1}$），而原函数取不到 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}$ 这一值（因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{1}$）。

通过解方程 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 反解出 $\htmlData{tutor-start=0,tutor-end=1}{x}$，并根据原函数的定义域确定根式的符号，最后互换 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 得到反函数表达式。

详细展开： 令 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{1}$。 移项得：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{1}$。 两边开平方：$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=12}{\pm }\sqrt{\htmlData{tutor-start=18,tutor-end=19}{y} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{1}}$。 解得：$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=10}{\pm }\sqrt{\htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{1}}$。 此时需要根据原函数的定义域 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}$ 来确定符号。 因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}$。 而 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{y} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=23}{\geqslant }\htmlData{tutor-start=23,tutor-end=24}{0}$，所以要使 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1}$ 为负数，必须取负号。 即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{y} \htmlData{tutor-start=17,tutor-end=18}{-} \htmlData{tutor-start=19,tutor-end=20}{1}}$，故 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{1}}$。 最后，将 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 互换，得到反函数解析式： $\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{1} \htmlData{tutor-start=14,tutor-end=15}{-} \sqrt{\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{1}}$。 结合第一步得出的定义域 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$，最终结果为： $\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{1} \htmlData{tutor-start=14,tutor-end=15}{-} \sqrt{\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{1}} \htmlData{tutor-start=29,tutor-end=31}{\ }\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{x} \htmlData{tutor-start=34,tutor-end=35}{>} \htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{)}$。 对比剩余选项 (A) 和 (B)： (A) 为 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{1}}$，对应 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$ 的情况，不符。 (B) 为 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{-} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{1}}$，对应 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$ 的情况，符合。 故选 (B)。

将不等式组转化为直线方程，画出可行域，并求出边界直线的交点（顶点）。

详细展开： 约束条件： 1. $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{y} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=20}{\geqslant }\htmlData{tutor-start=20,tutor-end=21}{0} \implies \htmlData{tutor-start=31,tutor-end=32}{y} \htmlData{tutor-start=33,tutor-end=43}{\leqslant }\htmlData{tutor-start=43,tutor-end=44}{x} \htmlData{tutor-start=45,tutor-end=46}{+} \htmlData{tutor-start=47,tutor-end=48}{1}$ 2. $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{y} \htmlData{tutor-start=6,tutor-end=16}{\geqslant }\htmlData{tutor-start=16,tutor-end=17}{0} \implies \htmlData{tutor-start=27,tutor-end=28}{y} \htmlData{tutor-start=29,tutor-end=39}{\geqslant }\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{x}$ 3. $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$

求边界交点： - 联立 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}$： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x} \implies \htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1} \implies \htmlData{tutor-start=37,tutor-end=38}{x} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{/}\htmlData{tutor-start=44,tutor-end=45}{2}$。 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}$。 交点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$。检查 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$，满足。

- 联立 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}$： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0} \implies \htmlData{tutor-start=15,tutor-end=16}{y} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}$。 交点 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}$。检查 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x} \implies \htmlData{tutor-start=24,tutor-end=25}{1} \htmlData{tutor-start=26,tutor-end=36}{\geqslant }\htmlData{tutor-start=36,tutor-end=37}{0}$，满足。

- 联立 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}$： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0} \implies \htmlData{tutor-start=15,tutor-end=16}{y} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{0}$。 交点 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$。检查 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1} \implies \htmlData{tutor-start=27,tutor-end=28}{0} \htmlData{tutor-start=29,tutor-end=39}{\leqslant }\htmlData{tutor-start=39,tutor-end=40}{1}$，满足。

可行域是由点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$, $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}$, $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 围成的三角形区域（包含边界）。

将各顶点坐标代入目标函数 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{y}$，找出最小值。

详细展开： 目标函数 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{y}$。 1. 在点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$ 处： $\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{A}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{2} \htmlData{tutor-start=29,tutor-end=30}{+} \htmlData{tutor-start=31,tutor-end=32}{1} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{/}\htmlData{tutor-start=37,tutor-end=38}{2}$。 2. 在点 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}$ 处： $\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{B}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{0} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{2}$。 3. 在点 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 处： $\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{C}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{0} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{0}$。

比较三个值：$\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2} \htmlData{tutor-start=8,tutor-end=9}{<} \htmlData{tutor-start=10,tutor-end=11}{2}$。 所以最小值为 0。

利用等差数列通项公式建立方程组，解出首项 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 和公差 $\htmlData{tutor-start=0,tutor-end=1}{d}$。

详细展开： 设等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的公差为 $\htmlData{tutor-start=0,tutor-end=1}{d}$。 由题意知： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{d} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{6}$ ① $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{d} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{5}$ ② 由 ② - ① 得： $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{d} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{9} \implies \htmlData{tutor-start=16,tutor-end=17}{d} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{3}$ 将 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 代入 ① 得： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{3} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{6} \implies \htmlData{tutor-start=23,tutor-end=24}{a}_{\htmlData{tutor-start=26,tutor-end=27}{1}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{3}$ 所以，$\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的通项公式为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{3} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{n}$。

根据 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{n}}$ 写出 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前几项或通项，判断其性质并求前 5 项和。

详细展开： 因为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{n}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{n}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{6}\htmlData{tutor-start=26,tutor-end=27}{n}$。 可以看出 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 也是一个等差数列，首项 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{6}$，公差 $\htmlData{tutor-start=0,tutor-end=1}{d}' \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{6}$。 我们需要求 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 5 项和 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}}$。 方法一：直接列出前 5 项 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{b}_{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{b}_{\htmlData{tutor-start=26,tutor-end=27}{3}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{8}\htmlData{tutor-start=33,tutor-end=34}{,} \htmlData{tutor-start=35,tutor-end=36}{b}_{\htmlData{tutor-start=38,tutor-end=39}{4}} \htmlData{tutor-start=41,tutor-end=42}{=} \htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{,} \htmlData{tutor-start=47,tutor-end=48}{b}_{\htmlData{tutor-start=50,tutor-end=51}{5}} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{3}\htmlData{tutor-start=56,tutor-end=57}{0}$ $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{6} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{8} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{4} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{0} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{9}\htmlData{tutor-start=33,tutor-end=34}{0}$ 方法二：使用等差数列求和公式 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{b}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{b}_{\htmlData{tutor-start=27,tutor-end=28}{5}}\htmlData{tutor-start=29,tutor-end=30}{)}}{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{=} \frac{\htmlData{tutor-start=43,tutor-end=44}{5}\htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{6} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{3}\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=52}{)}}{\htmlData{tutor-start=54,tutor-end=55}{2}} \htmlData{tutor-start=57,tutor-end=58}{=} \frac{\htmlData{tutor-start=65,tutor-end=66}{5} \htmlData{tutor-start=67,tutor-end=74}{\times }\htmlData{tutor-start=74,tutor-end=75}{3}\htmlData{tutor-start=75,tutor-end=76}{6}}{\htmlData{tutor-start=78,tutor-end=79}{2}} \htmlData{tutor-start=81,tutor-end=82}{=} \htmlData{tutor-start=83,tutor-end=84}{5} \htmlData{tutor-start=85,tutor-end=92}{\times }\htmlData{tutor-start=92,tutor-end=93}{1}\htmlData{tutor-start=93,tutor-end=94}{8} \htmlData{tutor-start=95,tutor-end=96}{=} \htmlData{tutor-start=97,tutor-end=98}{9}\htmlData{tutor-start=98,tutor-end=99}{0}$

为了量化分析，建立空间直角坐标系。设正方体棱长为 1。以 D 为原点，DA, DC, DD_1 所在直线分别为 x, y, z 轴。

详细展开： 1. 顶点坐标：D(0,0,0), A(1,0,0), B(1,1,0), C(0,1,0), D_1(0,0,1), B_1(1,1,1)。 2. 对角线 BD_1 的方向向量：\vec{BD_1} = D_1 - B = (-1, -1, 1)。其模长 |BD_1| = \sqrt{(-1)^2 + (-1)^2 + 1^2} = \sqrt{3}。 3. 平面 BB_1D_1D 即对角面，其法向量可以通过观察得出。平面 BB_1D_1D 包含 z 轴和直线 y=x (在 xy 平面投影)。实际上，该平面垂直于底面 ABCD 的对角线 AC。向量 \vec{AC} = (-1, 1, 0) 垂直于 \vec{DB}=(1,1,0) 且垂直于 z 轴，故 \vec{n} = (1, -1, 0) 是平面 BB_1D_1D 的一个法向量（或者取 (-1, 1, 0)）。 4. 题目条件：直线 MN 过点 P 且垂直于平面 BB_1D_1D。这意味着直线 MN 的方向向量平行于平面的法向量 \vec{n} = (1, -1, 0)。 5. 因此，直线 MN 平行于底面对角线 AC，且位于过点 P 的水平截面内（因为法向量 z 分量为 0，说明直线 MN 是水平的，即 z 坐标不变？不，法向量是 (1,-1,0)，直线方向也是 (1,-1,0)。这意味着直线 MN 上所有点的 z 坐标相同吗？是的，因为方向向量 z 分量为 0。所以 MN 是一条水平线段。） 6. 等等，让我们重新检查。平面 BB_1D_1D 的方程。B(1,1,0), D(0,0,0), D_1(0,0,1)。向量 \vec{DB}=(1,1,0), \vec{DD_1}=(0,0,1)。法向量 \vec{n} = \vec{DB} \times \vec{DD_1} = (1, -1, 0)。正确。 7. 直线 MN 垂直于平面，故其方向向量为 (1, -1, 0)。这意味着直线 MN 平行于 xOy 平面。也就是说，对于对角线 BD_1 上的任意一点 P，过 P 作垂直于对角面的直线，这条直线一定在过 P 且平行于底面的平面内吗？不一定。直线 MN 的方向是固定的 (1, -1, 0)。如果 P 的坐标是 (x_P, y_P, z_P)，则直线 MN 的参数方程为 x = x_P + t, y = y_P - t, z = z_P。由此可见，直线上所有点的 z 坐标均为 z_P。因此，MN 确实位于高度为 z_P 的水平面上。 8. 结论：MN 是正方体在高度 z = z_P 处的水平截面（正方形）中，平行于对角线 AC 的弦。

设正方体棱长为 a（为方便计算，后续可归一化，这里保留 a 以体现比例关系，最后发现形状与 a 无关）。对角线 BD_1 长为 \sqrt{3}a。BP = x。

详细展开： 1. 点 P 在 BD_1 上。B 点对应 x=0，D_1 点对应 x=\sqrt{3}a。 2. 利用相似比或向量分解。P 点在竖直方向的高度 z_P 与 BP 的长度 x 成正比。B 点 z=0，D_1 点 z=a。总长 L=\sqrt{3}a。所以 z_P = \frac{x}{\sqrt{3}a} \cdot a = \frac{x}{\sqrt{3}}。 3. 直线 MN 位于高度 z = z_P 的平面上。该平面截正方体得到一个边长为 a 的正方形 A'B'C'D'（实际上是 ABCD 向上平移 z_P）。 4. 在该正方形截面中，直线 MN 的方向平行于对角线 AC（即方向 (1, -1)）。直线过点 P'（P 在该截面上的投影，也就是 P 本身，因为 P 就在该高度）。 5. 我们需要确定 P' 在截面正方形中的位置。P 在对角线 BD_1 上，其在底面的投影 P_{xy} 位于底面对角线 BD 上。由于对称性，P_{xy} 到中心 O 的距离与高度有关，或者直接看截面内的几何结构。 6. 更简单的方法：考虑截面正方形内的几何性质。截面是边长为 a 的正方形。直线 MN 平行于一条对角线（如 AC），且经过点 P。点 P 在正方体体对角线 BD_1 上。注意，体对角线 BD_1 在水平面上的投影是底面对角线 BD。而在正方形 ABCD 中，BD \perp AC。所以在任何水平截面中，体对角线上的点 P 的投影 P' 位于截面正方形的一条对角线（对应 BD 方向）上。 7. 问题转化为：在边长为 a 的正方形中，有一条弦 MN 平行于对角线 AC，且经过另一条对角线 BD 上的点 P'。求弦长 MN。 8. 设正方形中心为 O。P' 到中心 O 的距离为 d。由几何关系，当 P' 在中心时（即正方体中心），弦长最长，为对角线长 \sqrt{2}a。当 P' 移向顶点 B 或 D 时，弦长变短。 9. 具体计算：设 P' 到最近的边（或顶点）的距离。令 P' 沿对角线 BD 移动。设 P' 到中心 O 的距离为 u。则弦长 MN = 2 \times (半对角线长 - u \times \cos(45^\circ) ? 不对)。 让我们用解析几何。在截面正方形内建立局部坐标系，中心为原点，对角线方向为轴。或者直接用相似三角形。 当 P 从 B 运动到 D_1 时，z 从 0 增加到 a。P 在底面投影从 B 移动到 D。不，P 在 BD_1 上，其水平投影 P_{xy} 从 B(1,1) 移动到 D(0,0)（假设 D 为原点，则 B 为 (a,a)，D 为 (0,0)，投影是从 (a,a) 到 (0,0)）。 等等，前面的坐标系是 D(0,0,0), B(1,1,0)。BD_1 连接 (1,1,0) 和 (0,0,1)。 P 点坐标：B + \frac{x}{\sqrt{3}} \vec{u}_{BD_1}。\vec{u}_{BD_1} = \frac{1}{\sqrt{3}}(-1, -1, 1)。 P(x_P, y_P, z_P) = (1, 1, 0) + \frac{x}{\sqrt{3}} \frac{1}{\sqrt{3}} (-1, -1, 1) = (1 - \frac{x}{3}, 1 - \frac{x}{3}, \frac{x}{3})。（这里假设 a=1）。 截面高度 z = \frac{x}{3}。截面正方形区域：x \in [0,1], y \in [0,1]。 直线 MN 过 P，方向 (1, -1)。方程：X - x_P = -(Y - y_P) \Rightarrow X + Y = x_P + y_P。 代入 P 点坐标：X + Y = (1 - \frac{x}{3}) + (1 - \frac{x}{3}) = 2 - \frac{2x}{3}。 直线 X + Y = k 截正方形 [0,1]\times[0,1] 的弦长。 当 k \in [1, 2] 时（对应 x 从 0 到 1.5? 不，x 范围是 0 到 \sqrt{3} \approx 1.732。k 的范围：x=0, k=2; x=\sqrt{3}, k=2-2\sqrt{3}/3 \approx 2-1.15=0.85。中间经过 k=1）。 正方形被直线 X+Y=k 截得的线段长度 L(k)： - 当 1 \le k \le 2 时，直线截去右上角三角形。交点为 (k-1, 1) 和 (1, k-1)。距离 = \sqrt{(1-(k-1))^2 + ((k-1)-1)^2} = \sqrt{2(2-k)^2} = \sqrt{2}(2-k)。 - 当 0 \le k < 1 时，直线截去左下角三角形。交点为 (k, 0) 和 (0, k)。距离 = \sqrt{k^2+k^2} = \sqrt{2}k。

现在将 k 换回 x。k = 2 - \frac{2x}{3}。 分界点 k=1 \Rightarrow 2 - \frac{2x}{3} = 1 \Rightarrow \frac{2x}{3} = 1 \Rightarrow x = 1.5。 注意正方体对角线长 \sqrt{3} \approx 1.732。1.5 < 1.732。所以存在分段。

第一段：0 \le x \le 1.5 (此时 1 \le k \le 2)。 y = \sqrt{2} (2 - (2 - \frac{2x}{3})) = \sqrt{2} (\frac{2x}{3}) = \frac{2\sqrt{2}}{3} x。 这是一个正比例函数，图象为过原点的直线段。

第二段：1.5 < x \le \sqrt{3} (此时 0.85 \le k < 1)。 y = \sqrt{2} (2 - \frac{2x}{3})。 这是一个一次函数，斜率为负。y 随 x 增大而减小。 当 x = 1.5 时，y = \sqrt{2}。当 x = \sqrt{3} 时，y = \sqrt{2}(2 - \frac{2\sqrt{3}}{3}) = 2\sqrt{2} - \frac{2\sqrt{6}}{3} > 0。

综上，函数 y=f(x) 先线性增加，达到最大值后线性减小。图象是一个折线，先升后降。 对比选项： A: 曲线，凸。 B: 折线，先升后降。 C: 曲线，凹。 D: 单调。

因此选 B。

根据任意角三角函数的定义，由角 \alpha 终边上一点 P(x, y) 的坐标直接确定 \tan \alpha 的值。

详细展开： 已知角 \alpha 的终边经过点 P(1, -2)。 根据正切函数的定义：\tan \alpha = \frac{y}{x}，其中 x \neq 0。 代入 x = 1, y = -2，得： \tan \alpha = \frac{-2}{1} = -2。

将求得的 \tan \alpha 代入正切的二倍角公式进行计算。

详细展开： 正切的二倍角公式为：\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^2 \alpha}。 将 \tan \alpha = -2 代入公式： \tan 2\alpha = \frac{2 \times (-2)}{1 - (-2)^2} = \frac{-4}{1 - 4} = \frac{-4}{-3} = \frac{4}{3}。

将不等式右边的常数 1 移至左边，通过通分合并为一个分式，从而避免直接去分母时因符号不确定而产生的错误。

详细展开： 原不等式为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{1}$。 首先，确定定义域：分母不能为零，即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2} \neq \htmlData{tutor-start=11,tutor-end=12}{0}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}$。 接着，将 1 移项至左边： $\frac{\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=25}{>} \htmlData{tutor-start=26,tutor-end=27}{0}$ 对左边进行通分，公分母为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}$： $\frac{\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}}{\htmlData{tutor-start=23,tutor-end=24}{x} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{>} \htmlData{tutor-start=32,tutor-end=33}{0}$ 化简分子： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{2} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{3}$ 因此，不等式转化为： $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{>} \htmlData{tutor-start=19,tutor-end=20}{0}$

根据分式的符号法则，结合分子为负常数的事实，推导分母的符号限制，进而得出 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的范围。

详细展开： 我们得到不等式 $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{>} \htmlData{tutor-start=19,tutor-end=20}{0}$。 由于分子 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{0}$ 恒成立，要使整个分式的值大于 0，分母 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}$ 必须与分子异号，即分母必须小于 0。 因此有： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}$ 解得： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ 等等，让我们重新检查计算。 分子是 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{x} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{2} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{3}$。是的，分子是 -3。 不等式是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{>} \htmlData{tutor-start=17,tutor-end=18}{0}$。 因为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}$ 是负数，所以 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}$ 必须是负数，才能使商为正。 即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0} \implies \htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{<} \htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{2}$。

让我再次仔细检查题目和常规解法，防止低级错误。 题目：$\frac{\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{1}$ 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}$ (小于-2): $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{3} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \frac{\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{4}}{\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{4} \htmlData{tutor-start=42,tutor-end=43}{>} \htmlData{tutor-start=44,tutor-end=45}{1}$. 成立。 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}$ (大于-2): $\frac{\htmlData{tutor-start=6,tutor-end=7}{0} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{0} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{5} \htmlData{tutor-start=27,tutor-end=28}{>} \htmlData{tutor-start=29,tutor-end=30}{1}$. 不成立。 取 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$ (大于-2): $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{2} \htmlData{tutor-start=27,tutor-end=28}{>} \htmlData{tutor-start=29,tutor-end=30}{1}$. 不成立。

看来我的初步推导 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ 是正确的？ 让我们再看一眼常见的这类题目陷阱。通常这类题目会有两个边界。难道我算错了？ 再算一遍通分： $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{1} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{0}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)}}{\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{>} \htmlData{tutor-start=26,tutor-end=27}{0}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{-} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{2}}{\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{>} \htmlData{tutor-start=28,tutor-end=29}{0}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{>} \htmlData{tutor-start=17,tutor-end=18}{0}$ 这等价于 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}$ 同号？不，商大于0，说明分子分母同号。 分子 -3 是负的。 所以分母 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}$ 必须是负的。 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2} \htmlData{tutor-start=4,tutor-end=5}{<} \htmlData{tutor-start=6,tutor-end=7}{0} \htmlData{tutor-start=8,tutor-end=23}{\implies x < -2}$。

但是，通常北京卷填空题第10题如果是这种简单线性分式，答案往往是开区间。如果是 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$，集合表示为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$。

让我反思一下是否有其他可能性。比如题目抄写错误？ 如果不等式是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0}$，那解集是 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$。 如果不等式是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{1}$，那 $\frac{-3}{x+2} < 0 \implies x+2 > 0 \implies x > -2$。

等等，我是否看错了题目中的数字？ 题目文本：`\frac{x - 1}{x + 2} > 1`

让我们尝试另一种方法验证：图像法或特值法。 函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{=} \frac{\htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{3}}{\htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{2}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{1} \htmlData{tutor-start=47,tutor-end=48}{-} \frac{\htmlData{tutor-start=55,tutor-end=56}{3}}{\htmlData{tutor-start=58,tutor-end=59}{x}\htmlData{tutor-start=59,tutor-end=60}{+}\htmlData{tutor-start=60,tutor-end=61}{2}}$。 我们要解 $\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{-} \frac{\htmlData{tutor-start=10,tutor-end=11}{3}}{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{>} \htmlData{tutor-start=20,tutor-end=21}{1}$。 两边减 1： $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{>} \htmlData{tutor-start=17,tutor-end=18}{0}$ 两边乘 -1，变号： $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{<} \htmlData{tutor-start=16,tutor-end=17}{0}$ 因为分子 3 > 0，所以分母必须 < 0。 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0} \implies \htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{<} \htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{2}$。

结论确实是 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$。

但是，作为AI，我必须非常谨慎。有没有可能题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}$ 或者其他？不，用户给的是 `> 1`。 有没有可能我看错了已知答案？用户给的 `knownAnswer` 是空的。

让我再检查一种情况，是不是我把减法做错了？ $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{3}$。没错。

好吧，如果答案真的是 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$，那就这样写。但通常这类“填空第10题”在北京卷中难度适中，有时候会考察更复杂的区间，比如 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)}$ 形式。例如，如果题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{1}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$ 之类的限制？没有。

或者，题目其实是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0}$？不，我要忠实于用户输入。 用户输入：`\frac{x - 1}{x + 2} > 1`

再仔细看一眼。有没有可能题目是 $\left| \frac{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}} \right| \htmlData{tutor-start=31,tutor-end=32}{>} \htmlData{tutor-start=33,tutor-end=34}{1}$？没有绝对值符号。

那就按 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ 输出。解集为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$。

*自我修正*：等一下，我再次模拟一下考试常见题型。很多时候题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{1}$ 或者类似。但这里是严格大于。

让我们再试一个值。$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{0}$。 LHS: $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{8} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{.}\htmlData{tutor-start=29,tutor-end=30}{3}\htmlData{tutor-start=30,tutor-end=31}{7}\htmlData{tutor-start=31,tutor-end=32}{5}$。 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{7}\htmlData{tutor-start=4,tutor-end=5}{5} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{1}$。成立。 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}$。 LHS: $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{4}$。 $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$。成立。 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{1}$。 LHS: $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{.}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{.}\htmlData{tutor-start=28,tutor-end=29}{1} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{1}$。 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{1}$。成立。 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{9}$。 LHS: $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{.}\htmlData{tutor-start=23,tutor-end=24}{9}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{.}\htmlData{tutor-start=27,tutor-end=28}{1} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{9}$。 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{9} \htmlData{tutor-start=4,tutor-end=5}{>} \htmlData{tutor-start=6,tutor-end=7}{1}$。不成立。

所以解集确实是 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$。

但是，为了保险起见，我将再次核对是否有可能题目是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}$? 如果是 -1: $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{1} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{0} \implies \frac{\htmlData{tutor-start=39,tutor-end=40}{x}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{2}}{\htmlData{tutor-start=48,tutor-end=49}{x}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{2}} \htmlData{tutor-start=53,tutor-end=54}{>} \htmlData{tutor-start=55,tutor-end=56}{0} \implies \frac{\htmlData{tutor-start=72,tutor-end=73}{2}\htmlData{tutor-start=73,tutor-end=74}{x}\htmlData{tutor-start=74,tutor-end=75}{+}\htmlData{tutor-start=75,tutor-end=76}{1}}{\htmlData{tutor-start=78,tutor-end=79}{x}\htmlData{tutor-start=79,tutor-end=80}{+}\htmlData{tutor-start=80,tutor-end=81}{2}} \htmlData{tutor-start=83,tutor-end=84}{>} \htmlData{tutor-start=85,tutor-end=86}{0}$。 根为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}$。大于取两边：$\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}$ 或 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$。

如果是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{1}$? $\frac{-3}{x+2} < 0 \implies x+2 > 0 \implies x > -2$。

鉴于用户提供的题目明确是 `> 1`，我将提供 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$ 作为答案。这在数学上是完全正确的。

*再次检查*：有没有可能题目中的分母是 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}$？ 如果是 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{1} \implies \frac{\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{x}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{)}}{\htmlData{tutor-start=46,tutor-end=47}{x}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=52}{>} \htmlData{tutor-start=53,tutor-end=54}{0} \implies \frac{\htmlData{tutor-start=70,tutor-end=71}{1}}{\htmlData{tutor-start=73,tutor-end=74}{x}\htmlData{tutor-start=74,tutor-end=75}{-}\htmlData{tutor-start=75,tutor-end=76}{2}} \htmlData{tutor-start=78,tutor-end=79}{>} \htmlData{tutor-start=80,tutor-end=81}{0} \implies \htmlData{tutor-start=91,tutor-end=92}{x} \htmlData{tutor-start=93,tutor-end=94}{>} \htmlData{tutor-start=95,tutor-end=96}{2}$。

不管怎样，基于输入文本 `x + 2`，答案是 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}$。

明确向量数量积（点积）的计算公式，识别题目中给出的已知量。

详细展开： 向量数量积的定义公式为： $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|} \htmlData{tutor-start=55,tutor-end=56}{|}\boldsymbol{\htmlData{tutor-start=68,tutor-end=69}{b}}\htmlData{tutor-start=70,tutor-end=71}{|} \cos \htmlData{tutor-start=77,tutor-end=83}{\theta}$ 其中，$\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}$ 分别是向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 和 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}$ 的模（长度），$\htmlData{tutor-start=0,tutor-end=6}{\theta}$ 是这两个向量的夹角。

从题设中提取已知条件： 1. 向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 的模 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}$ 2. 向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}$ 的模 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}$ 3. 向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}$ 与 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}$ 的夹角 $\theta = 120^\circ$

将已知数值代入公式，计算特殊角的三角函数值，得出最终结果。

详细展开： 将 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}$, $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}$, $\theta = 120^\circ$ 代入公式： $\boldsymbol{a} \cdot \boldsymbol{b} = 4 \times 4 \times \cos 120^\circ$

接下来计算 $\cos 120^\circ$ 的值。 根据诱导公式 $\cos(180^\circ - \alpha) = -\cos \alpha$，或者单位圆知识： $\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ$ 我们知道 $\cos 60^\circ = \frac{1}{2}$， 所以 $\cos 120^\circ = -\frac{1}{2}$。

继续计算： $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{6} \htmlData{tutor-start=41,tutor-end=48}{\times }\left( \htmlData{tutor-start=55,tutor-end=56}{-}\frac{\htmlData{tutor-start=62,tutor-end=63}{1}}{\htmlData{tutor-start=65,tutor-end=66}{2}} \right)$ $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{8}$

因此，$\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}$ 的值为 -8。

写出二项展开式的通项公式 T_{r+1}，令 x 的指数为 0 求出 r，进而求出常数项。

详细展开： 1. 二项式为 \left(x^2 + \frac{1}{x^3}\right)^5，即 (x^2 + x^{-3})^5。 2. 通项公式：T_{r+1} = C_5^r (x^2)^{5-r} (x^{-3})^r。 3. 整理 x 的指数：T_{r+1} = C_5^r x^{2(5-r)} x^{-3r} = C_5^r x^{10 - 2r - 3r} = C_5^r x^{10 - 5r}。 4. 令指数为 0：10 - 5r = 0 \Rightarrow 5r = 10 \Rightarrow r = 2。 5. 代入 r = 2 求系数：C_5^2 = \frac{5 \times 4}{2 \times 1} = 10。 6. 所以常数项为 10。

利用赋值法，令变量 x = 1，即可得到各项系数之和。

详细展开： 1. 设 f(x) = \left(x^2 + \frac{1}{x^3}\right)^5 = \sum a_i x^{k_i}。 注意：这里的“各项系数之和”通常指二项式展开后，每一项前面的数字因子之和。对于形如 (ax^m + bx^n)^N 的展开式，令 x=1 即可求得所有项的系数代数和。 2. 令 x = 1。 3. 原式变为 \left(1^2 + \frac{1}{1^3}\right)^5 = (1 + 1)^5 = 2^5。 4. 计算 2^5 = 32。 5. 所以各项系数之和为 32。

先从端点 A 读取 f(0)，再把所得数作为新的自变量。

详细展开：图象给出 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}$。在 BC 段上斜率为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}$，故 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}$，于是 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}$。