2014 年高考数学（新课标卷 2文科）

怎么想到：分母含虚数单位时，乘分母的共轭复数可把分母化为实数，这是复数除法的标准入口。

为什么从这里入手：复数题先判断目标需要代数形式还是模与辐角。“怎么想到：分母含虚数单位时，乘分母的共轭复数可把分母化为实数，这是复数除法的标准入口。”与“识别结构并建立关系”直接相连，先利用共轭、模或实虚部关系，通常能避免把复数完全展开成冗长乘积。

详细展开：分母$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$的共轭复数是$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$。分子、分母同乘$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}$不改变商，分母成为$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{+}\mathrm{\htmlData{tutor-start=25,tutor-end=26}{i}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{-}\mathrm{\htmlData{tutor-start=39,tutor-end=40}{i}}^{\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{2}$。分子展开为$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\mathrm{\htmlData{tutor-start=22,tutor-end=23}{i}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{3}\mathrm{\htmlData{tutor-start=34,tutor-end=35}{i}}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{4}\mathrm{\htmlData{tutor-start=53,tutor-end=54}{i}}$，再除以2。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：反向核验：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{-}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{3}\mathrm{\htmlData{tutor-start=42,tutor-end=43}{i}}$，恰好还原原分子，所以商正确，选择B。

怎么想到：充分性要问“导数为零是否一定有极值”，必要性要问“可导且有极值是否一定导数为零”；两个方向必须分别检验。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：充分性要问“导数为零是否一定有极值”，必要性要问“可导且有极值是否一定导数为零”；两个方向必须分别检验。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：若$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$是可导函数的极值点，由费马定理必有$\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{0}}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}$，所以$\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=13}{\Rightarrow }\htmlData{tutor-start=13,tutor-end=14}{p}$，p是q的必要条件。反过来，取$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{3}}$且$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}$，虽然$\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}$，但$\htmlData{tutor-start=0,tutor-end=1}{f}$在0两侧都递增，0不是极值点，所以$\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=14}{\nRightarrow }\htmlData{tutor-start=14,tutor-end=15}{q}$，p不是充分条件。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：必要性由可导极值点定理保证，反例$\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}}$又确定性地否定充分性。因此唯一正确表述是“必要但不充分”，选择C。

怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$与$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$展开后只有数量积项符号相反，作差可直接隔离$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=7}{\cdot }\htmlData{tutor-start=7,tutor-end=8}{b}$。

为什么从这里入手：代数题要寻找能减少未知量或降低次数的关系。“怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$与$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$展开后只有数量积项符号相反，作差可直接隔离$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=7}{\cdot }\htmlData{tutor-start=7,tutor-end=8}{b}$。”正好提供了因式、根或参数之间的等式，所以先识别结构并建立关系，再代入、消元或比较系数，计算链条会更短也更容易复核。

详细展开：平方展开得到$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{|}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=34}{\cdot }\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{0}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{|}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{|}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=34}{\cdot }\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{6}$。第一式减第二式，公共的两个模平方消去，得到$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}$可得$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=7}{\cdot }\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}$。把1代回两展开式，两式右侧之差确为4，与已知$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{6}$一致，选择A。

怎么想到：三个已知下标项成等比数列意味着中间项平方等于两端项乘积；公差已知后，所有项都可用$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$表示。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“怎么想到：三个已知下标项成等比数列意味着中间项平方等于两端项乘积；公差已知后，所有项都可用$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$表示。”提示了差、比或可累加关系，先识别结构并建立关系能把第 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：公差$\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$，所以$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}$，$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{6}$，$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{8}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{4}$。成等比数列给出$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{8}}$。展开后$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{)}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{4}\htmlData{tutor-start=32,tutor-end=33}{)}$，消去$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}$得$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{6}\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{8}$，故$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}$。于是$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{n}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：核验条件：$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{8}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{6}$，相邻比都为2，确成等比数列；前n项$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{+}\cdots\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=21}{\cdot }\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{n}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{n}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{)}$，选择A。

怎么想到：原卷三视图中端面是半径3与半径2的同心圆，两个轴向视图显示长度4处半径2、长度2处半径3，因此零件应按两段同轴圆柱相加，而不是把内圆误判为通孔。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：原卷三视图中端面是半径3与半径2的同心圆，两个轴向视图显示长度4处半径2、长度2处半径3，因此零件应按两段同轴圆柱相加，而不是把内圆误判为通孔。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：原毛坯是半径3、高6的圆柱，体积$\htmlData{tutor-start=0,tutor-end=1}{V}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=9}{\pi}\htmlData{tutor-start=9,tutor-end=14}{\cdot}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\htmlData{tutor-start=24,tutor-end=25}{6}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=31}{\pi}$。由网格逐格读出，零件前4 cm的一段半径为2，后2 cm的一段半径为3，所以保留体积$\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{2}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=20}{\cdot}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=25}{\pi}\htmlData{tutor-start=25,tutor-end=30}{\cdot}\htmlData{tutor-start=30,tutor-end=31}{3}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=40}{\cdot}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{6}\htmlData{tutor-start=44,tutor-end=47}{\pi}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{8}\htmlData{tutor-start=50,tutor-end=53}{\pi}\htmlData{tutor-start=53,tutor-end=54}{=}\htmlData{tutor-start=54,tutor-end=55}{3}\htmlData{tutor-start=55,tutor-end=56}{4}\htmlData{tutor-start=56,tutor-end=59}{\pi}$。切削体积为$\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=17}{\pi}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：尺寸核验：两段轴向长度$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{6}$等于毛坯高度，最大半径3不超过毛坯半径；保留比例$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{7}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{7}$与切削比例$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{7}$相加为1，故选择C。

怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{1}}$都在与$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$同向的同一侧面内，以该三角形作底面，顶点A到侧面的距离正好是正三角形ABC的高。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{1}}$都在与$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$同向的同一侧面内，以该三角形作底面，顶点A到侧面的距离正好是正三角形ABC的高。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：底面ABC为边长2的正三角形，D是BC中点，所以$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}}$且$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C}$。侧面$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{C}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}}$是长2、宽$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$的矩形。三角形$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{D}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}$以$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{2}$为底，D到直线$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}$的距离等于侧棱长$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$，故面积为$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=16}{\cdot}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=22}{\cdot}\sqrt{\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{=}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{3}}$。点A到该侧面的距离是$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：用坐标复核时可取$\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{)}$、$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{)}$，行列式绝对值除以6仍为1，故选择C。

怎么想到：程序框图必须严格按箭头执行；判断条件是$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{t}$，所以$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}$这一轮仍要计算，不能提前退出。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：程序框图必须严格按箭头执行；判断条件是$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{t}$，所以$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}$这一轮仍要计算，不能提前退出。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：输入$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$后初始化$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}$。第一轮$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}$，执行$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=12}{\cdot}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}$，$\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}$，再令$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$。第二轮$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}$仍成立，执行$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=12}{\cdot}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}$，$\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{7}$，再令$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$。此时$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}$不成立，输出S。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：核验循环次数：$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}$两次都满足不等式，共给S增加$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{4}$；初值3加4等于7，故选择D。

怎么想到：线性目标函数在封闭多边形可行域上的最值一定在顶点取得，因此先把三个不等式边界两两联立即可。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“怎么想到：线性目标函数在封闭多边形可行域上的最值一定在顶点取得，因此先把三个不等式边界两两联立即可。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：三个边界为$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}$、$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}$、$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}$。前两式交于$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$；第一、三式交于$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}$；第二、三式交于$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$。这三个点均满足全部不等式，构成可行域顶点。代入$\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{y}$，依次得到1、2、7，最大者为7。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：目标直线$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{z}$沿增大方向平移时，最后接触可行域的点是$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$；三个顶点值已经全部比较，故最大值7，选择B。

怎么想到：直线过抛物线焦点且给出倾斜角，这是焦点弦的典型信号；对$\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}$可使用$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{/}\sin^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=22}{\alpha}$。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：直线过抛物线焦点且给出倾斜角，这是焦点弦的典型信号；对$\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}$可使用$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{/}\sin^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=22}{\alpha}$。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：由$\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{x}$与标准式$\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}$比较，得到$\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{3}$。焦点弦与x轴的夹角为$\alpha=30^\circ$，所以$\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=14}{\alpha}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{4}$。代入公式，$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{/}\sin^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=22}{\alpha}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：也可把直线$\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{3}}$代入抛物线，求两交点后用距离公式，所得弦长同为12，故选择C。

怎么想到：含参数函数的单调性应先求导；要让整个区间都递增，导数必须对每个$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$非负。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：含参数函数的单调性应先求导；要让整个区间都递增，导数必须对每个$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$非负。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：$\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{x}$。当$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$时，$\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{1}$，且$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{x}$在$\htmlData{tutor-start=0,tutor-end=1}{x}$趋近1的过程中可以任意接近1。要使$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=8}{\ge}\htmlData{tutor-start=8,tutor-end=9}{0}$对所有$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$成立，必须且只需$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}$。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，$\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{>}\htmlData{tutor-start=12,tutor-end=13}{0}$，仍严格递增。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：若$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}$，可取足够接近1的$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}$使$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{k}$，此时导数为负，条件失败；若$\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}$则导数恒正。因此范围完整，选择D。

怎么想到：固定M后，射线MN绕MO转动；能与圆相交的最大偏角出现在射线与圆相切时，因此比较45度与切线半角。

为什么从这里入手：函数的局部变化由导数控制。“怎么想到：固定M后，射线MN绕MO转动；能与圆相交的最大偏角出现在射线与圆相切时，因此比较45度与切线半角。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：记$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{=}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{x}_{\htmlData{tutor-start=14,tutor-end=15}{0}}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}}\htmlData{tutor-start=23,tutor-end=26}{\ge}\htmlData{tutor-start=26,tutor-end=27}{1}$。若M在圆外，作切线MT，直角三角形OMT中$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{T}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}$，最大偏角$\htmlData{tutor-start=0,tutor-end=6}{\alpha}$满足$\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{T}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{O}\htmlData{tutor-start=15,tutor-end=16}{M}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{r}$。存在偏角45度的交线等价于$45^\circ\le\alpha$，即$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=13}{\le}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{r}$，从而$\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=4}{\le}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}$。代入$\htmlData{tutor-start=0,tutor-end=1}{r}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{0}}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}$得$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{1}$。M在圆上的$\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}$也可直接取合适弦，包含在此范围内。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：端点$\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}$时射线恰与圆相切，角为45度，可以取到；范围外最大偏角小于45度，不存在N。因此闭区间[-1,1]准确，选择A。

怎么想到：两人各有3种等可能选择，结果用有序对表示共有$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=7}{\times}\htmlData{tutor-start=7,tutor-end=8}{3}$种；再数同色的有序对。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：两人各有3种等可能选择，结果用有序对表示共有$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=7}{\times}\htmlData{tutor-start=7,tutor-end=8}{3}$种；再数同色的有序对。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：把红、白、蓝分别记为R、W、B。甲、乙的有序选择共有9种且等可能。同色事件只有$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{R}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{R}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{W}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{W}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{B}\htmlData{tutor-start=16,tutor-end=17}{)}$三种，所以所求概率为有利结果数3除以全部结果数9。也可先固定甲的颜色，乙与甲相同的概率始终是1/3。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：互补事件“不同色”有$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=7}{\times}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{6}$种，概率$\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{9}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}$，与$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$相加为1，核验无误。

怎么想到：式中同时出现$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=14}{\varphi}\htmlData{tutor-start=14,tutor-end=15}{)}$和$\cos \htmlData{tutor-start=5,tutor-end=6}{x}$，先展开前者，把$\sin \htmlData{tutor-start=5,tutor-end=6}{x}$、$\cos \htmlData{tutor-start=5,tutor-end=6}{x}$同类项合并，通常会还原成一个差角正弦。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“怎么想到：式中同时出现$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=14}{\varphi}\htmlData{tutor-start=14,tutor-end=15}{)}$和$\cos \htmlData{tutor-start=5,tutor-end=6}{x}$，先展开前者，把$\sin \htmlData{tutor-start=5,tutor-end=6}{x}$、$\cos \htmlData{tutor-start=5,tutor-end=6}{x}$同类项合并，通常会还原成一个差角正弦。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：利用$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=14}{\varphi}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{=}\sin \htmlData{tutor-start=21,tutor-end=22}{x}\cos\htmlData{tutor-start=26,tutor-end=33}{\varphi}\htmlData{tutor-start=33,tutor-end=34}{+}\cos \htmlData{tutor-start=39,tutor-end=40}{x}\sin\htmlData{tutor-start=44,tutor-end=51}{\varphi}$，原式变为$\sin \htmlData{tutor-start=5,tutor-end=6}{x}\cos\htmlData{tutor-start=10,tutor-end=17}{\varphi}\htmlData{tutor-start=17,tutor-end=18}{-}\cos \htmlData{tutor-start=23,tutor-end=24}{x}\sin\htmlData{tutor-start=28,tutor-end=35}{\varphi}$，这正是$\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=14}{\varphi}\htmlData{tutor-start=14,tutor-end=15}{)}$。无论参数$\htmlData{tutor-start=0,tutor-end=7}{\varphi}$为何值，x变化时正弦函数都能取到1。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：取$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=9}{\varphi}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=13}{\pi}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{k}\htmlData{tutor-start=18,tutor-end=21}{\pi}$时$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}$，上界可以达到，因此最大值不是仅有估计而是确切等于1。

怎么想到：偶函数给出关于y轴的对称$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}$；关于$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$对称又给出$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{h}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{)}$，应寻找从-1到3的最短映射链。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：偶函数给出关于y轴的对称$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}$；关于$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$对称又给出$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{h}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{)}$，应寻找从-1到3的最短映射链。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由偶函数性质，$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}$。点1和点3关于直线$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$对称，因为它们到2的距离都为1，所以由图象关于$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$对称可得$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{)}$。题设$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{3}$，故$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：第一步只用了偶对称，第二步只用了轴$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$对称，两个对应自变量都精确匹配，最终值与已知$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{)}$一致。

怎么想到：递推是分式变换，直接倒推七次容易出错；先把变换连续作用三次，检查是否回到原值。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“怎么想到：递推是分式变换，直接倒推七次容易出错；先把变换连续作用三次，检查是否回到原值。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：记$\htmlData{tutor-start=0,tutor-end=1}{T}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{)}$。计算$\htmlData{tutor-start=0,tutor-end=1}{T}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{[}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{]}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{x}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{/}\htmlData{tutor-start=29,tutor-end=30}{x}$，再算$\htmlData{tutor-start=0,tutor-end=1}{T}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{[}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{]}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{x}$，所以数列每三项循环。由$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{8}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{T}^{\htmlData{tutor-start=9,tutor-end=10}{7}}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{T}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{)}$，题设$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{8}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}$给出$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}$，解得$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：正向代入核验：$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{4}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{/}\htmlData{tutor-start=35,tutor-end=36}{2}$，三项循环，因此$\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{8}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}$，与题设一致。

怎么想到：条件$A+C=180^\circ$使$\cos \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\cos \htmlData{tutor-start=13,tutor-end=14}{A}$；对角线BD同时属于三角形ABD和BCD，两次余弦定理可消去BD。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“怎么想到：条件$A+C=180^\circ$使$\cos \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\cos \htmlData{tutor-start=13,tutor-end=14}{A}$；对角线BD同时属于三角形ABD和BCD，两次余弦定理可消去BD。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：设$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=8}{\alpha}$，则$C=180^\circ-\alpha$，所以$\cos \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\cos\htmlData{tutor-start=12,tutor-end=18}{\alpha}$。在三角形ABD中，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{D}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{B}\htmlData{tutor-start=24,tutor-end=30}{\cdot }\htmlData{tutor-start=30,tutor-end=31}{A}\htmlData{tutor-start=31,tutor-end=32}{D}\cos \htmlData{tutor-start=37,tutor-end=38}{A}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{4}\cos\htmlData{tutor-start=46,tutor-end=52}{\alpha}$。在三角形BCD中，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{C}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{C}\htmlData{tutor-start=15,tutor-end=16}{D}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{B}\htmlData{tutor-start=23,tutor-end=24}{C}\htmlData{tutor-start=24,tutor-end=30}{\cdot }\htmlData{tutor-start=30,tutor-end=31}{C}\htmlData{tutor-start=31,tutor-end=32}{D}\cos \htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{3}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}\cos\htmlData{tutor-start=48,tutor-end=54}{\alpha}$。两式相等得$\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}$，故$A=120^\circ,C=60^\circ$，再代回得$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{7}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：把$C=60^\circ$代入三角形BCD：$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=25}{\cdot}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=31}{\cdot}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=37}{\cdot}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{/}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{7}$，与三角形ABD所得相同，角和边同时核验。

怎么想到：四条边和角A、C都已知，分别用两边及夹角公式求$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{D}$与$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{D}$面积最直接。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：四条边和角A、C都已知，分别用两边及夹角公式求$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{D}$与$\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{D}$面积最直接。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：$S_{ABD}=\frac{1}{2} AB\cdot AD\sin A=\frac{1}{2}\cdot1\cdot2\cdot\sin120^\circ=\sqrt{3}/2$。同理$S_{BCD}=\frac{1}{2} BC\cdot CD\sin C=\frac{1}{2}\cdot3\cdot2\cdot\sin60^\circ=3\sqrt{3}/2$。两部分没有重叠且恰好拼成四边形。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两三角形共用对角线BD，位于其两侧；面积均为正且相加得到$\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}$，与边长量级相符。

怎么想到：E是PD中点，而矩形对角线AC、BD互相平分；取AC与BD交点O，就能在三角形PDB中使用中位线。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：E是PD中点，而矩形对角线AC、BD互相平分；取AC与BD交点O，就能在三角形PDB中使用中位线。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：设$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=9}{\cap }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{D}$。矩形的对角线互相平分，所以O是BD中点；题设E是PD中点。在三角形PDB中，连接两个中点E、O，得$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{B}$。又因$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{O}$都在平面AEC内，$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=9}{\subset}$平面AEC；PB不在该平面内，因此由线面平行判定定理，$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=11}{\parallel}$平面AEC。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：平行线EO确实完全位于目标平面中，且PB经过平面外点P，不会与EO重合，线面平行判定条件全部满足。

怎么想到：已知三棱锥P-ABD体积可先反求矩形边AB；PA垂直底面，建立直角坐标系后平面方程和点面距离都可直接算。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：已知三棱锥P-ABD体积可先反求矩形边AB；PA垂直底面，建立直角坐标系后平面方程和点面距离都可直接算。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{s}$。三角形ABD是矩形的一半，面积为$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{s}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}$，高$\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}$，故$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{V}_{\htmlData{tutor-start=14,tutor-end=15}{P}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{=}\frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{3}}\htmlData{tutor-start=32,tutor-end=37}{\cdot}\frac{\htmlData{tutor-start=43,tutor-end=44}{1}}{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{s}\sqrt{\htmlData{tutor-start=55,tutor-end=56}{3}}\htmlData{tutor-start=57,tutor-end=62}{\cdot}\htmlData{tutor-start=62,tutor-end=63}{1}$，解得$\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。取$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{D}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{,}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{3}}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{C}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{/}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{,}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{3}}\htmlData{tutor-start=54,tutor-end=55}{,}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{)}\htmlData{tutor-start=57,tutor-end=58}{,}\htmlData{tutor-start=58,tutor-end=59}{P}\htmlData{tutor-start=59,tutor-end=60}{=}\htmlData{tutor-start=60,tutor-end=61}{(}\htmlData{tutor-start=61,tutor-end=62}{0}\htmlData{tutor-start=62,tutor-end=63}{,}\htmlData{tutor-start=63,tutor-end=64}{0}\htmlData{tutor-start=64,tutor-end=65}{,}\htmlData{tutor-start=65,tutor-end=66}{1}\htmlData{tutor-start=66,tutor-end=67}{)}$。由P、B、C三点可得平面PBC方程$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\frac{\htmlData{tutor-start=8,tutor-end=9}{3}}{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{z}\htmlData{tutor-start=14,tutor-end=15}{-}\frac{\htmlData{tutor-start=21,tutor-end=22}{3}}{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}$。点A到该平面的距离为$\htmlData{tutor-start=0,tutor-end=1}{|}{\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{/}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{/}\sqrt{\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{3}}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：把B、C、P分别代入平面方程均得0；A代入的绝对值为$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}$。法向量长度为$\sqrt{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{2}$，所以距离$\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{3}}$计算一致。

怎么想到：样本量都是50，中位数不是某一个叶，而是从小到大第25项与第26项的平均数；先用各茎累计频数定位。

为什么从这里入手：统计量都来自明确的样本计数或加权。“怎么想到：样本量都是50，中位数不是某一个叶，而是从小到大第25项与第26项的平均数；先用各茎累计频数定位。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：甲部门60分及以下共有15个数据，所以第25、26项是70分段内第10、11个升序数据，分别为74、75，中位数$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{7}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{7}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{7}\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{5}$。乙部门50分及以下共有18个数据，所以第25、26项是60分段内第7、8个数据，分别为66、68，中位数$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{8}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{7}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两次定位都使用第25、26个观测值，且累计数量分别跨入甲的70分段和乙的60分段，与原茎叶图的中心位置一致。

怎么想到：题目要求“高于90”而不是“不低于90”，叶为0的90分必须排除；100分则计入。

为什么从这里入手：目标是“识别结构并建立关系”，而“怎么想到：题目要求“高于90”而不是“不低于90”，叶为0的90分必须排除；100分则计入。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：甲部门90分段为90、92、92、92、93、96，其中严格高于90的有5个，故估计概率$\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}$。乙部门90分段中严格高于90的有91、91、94、95、96共5个，100分段还有3个，总计8个，故估计概率$\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{6}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：逐项排除了恰等于90的数据并纳入三个100分，计数口径与“>90”完全一致；两概率都在[0,1]内。

怎么想到：评价分布不能只看一个高分比例；应同时比较中位数代表的中心位置、数据集中程度以及高分尾部。

为什么从这里入手：概率题的关键不是立刻代数，而是先说清随机试验与事件。“怎么想到：评价分布不能只看一个高分比例；应同时比较中位数代表的中心位置、数据集中程度以及高分尾部。”确定了基本事件或随机变量，所以先识别结构并建立关系，再依据独立、互斥、对立或期望线性性计算。

详细展开：甲部门中位数74.5高于乙部门67，且多数叶集中在60至80分，说明典型市民对甲的评分更高、意见更集中。乙部门从30多分延伸到100分，跨度更大，说明评价分歧更明显；但乙部门高于90分的样本比例0.16高于甲的0.10，表明乙也拥有更多极高评价。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：结论分别对应中心、尾部和离散性三项可观察证据，没有把“高分更多”误写成“总体一定更高”。因此可概括为甲总体较高且稳定，乙差异较大但高分尾部更多。

怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{2}}$与x轴垂直意味着M的横坐标就是右焦点横坐标c；把$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}$代入椭圆可得到M的纵坐标。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{2}}$与x轴垂直意味着M的横坐标就是右焦点横坐标c；把$\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}$代入椭圆可得到M的纵坐标。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：取上方交点$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{b}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{)}$，左焦点$\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}$。因为M、N、F1共线，直线MN的斜率就是MF1的斜率，即$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}$。令$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}$并用$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{c}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{a}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{e}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{)}$，得到$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{e}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{e}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{4}$。解$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{e}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{e}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}$，在$\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{e}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$内取$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$。

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：另一根$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}$不满足椭圆离心率$\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{e}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$；把$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$代回得$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}$，斜率条件完全复原。

怎么想到：焦点F1位于弦MN内部，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{F}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{N}$；给出的5倍关系能确定F1把弦分成4:1，从而由M坐标直接写N坐标。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“怎么想到：焦点F1位于弦MN内部，$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{F}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{N}$；给出的5倍关系能确定F1把弦分成4:1，从而由M坐标直接写N坐标。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：由$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{N}$得$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{F}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{N}$。向量方向相反，所以$\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{M}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{F}_{\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{c}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{b}^{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{/}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{)}$。直线MF1的y轴截距为$\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{b}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{]}\htmlData{tutor-start=13,tutor-end=14}{c}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{2}$，故$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{a}$。N在椭圆上，代入得$\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{c}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{6}\htmlData{tutor-start=25,tutor-end=26}{a}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{1}$。令$\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}$且$\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{e}^{\htmlData{tutor-start=17,tutor-end=18}{2}}$，化为$\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{e}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{e}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}$，解得$\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{7}$。于是$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{e}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{7}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{a}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{/}\htmlData{tutor-start=26,tutor-end=27}{a}$，故$\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{8}$。