题目 已知曲线 $C:xy=1$，过 $C$ 上一点 $A_{1}(x_{1},y_{1})$ 作斜率 $k_{1}$ 的直线，交曲线 $C$ 于另一点 $A_{2}\left(x_{2},y_{2}\right)$，再过 $A_{2}\left(x_{2},y_{2}\right)$ 作斜率为 $k_{2}$ 的直线，交曲线 $C$ 于另一点 $A_{3}(x_{3},y_{3})$，$\cdots$，过 $A_{n}\left(x_{n},y_{n}\right)$ 作斜率为 $k_{n}$ 的直线，交曲线 $C$ 于另一点 $A_{n+1}(x_{n+1},y_{n+1})$，$\cdots$，其中 $x_{1}=1$，$k_{n}=-\dfrac{x_{n}+1}{x_{n}^{2}+4x_{n}},x\in\mathbb N^{*}$． 问题1 求 $x_{n+1}$ 与 $x_{n}$ 的关系式； 答案1 $x_{n+1}=\dfrac{x_{n}+4}{x_{n}+1},n\in\mathbb N^{*}$ 解析1 由已知过 $A_{n}\left(x_{n},y_{n}\right)$ 斜率为 $-\dfrac{x_{n}+1}{x_{n}^{2}+4x_{n}}$ 的直线为$$y-y_{n}=-\dfrac{x_{n}+1}{x_{n}^{2}+4x_{n}}(x-x_{n}),$$直线交曲线 $C$ 于另一点 $A_{n+1}\left(x_{n+1},y_{n+1}\right)$，所以$$y_{n+1}-y_{n}=-\dfrac{x_{n}+1}{x_{n}^{2}+4x_{n}}(x_{n+1}-x_{n}),$$即\[\dfrac{1}{x_{n+1}}-\dfrac{1}{x_{n}}=-\dfrac{x_{n}+1}{x_{n}^{2}+4x_{n}}(x_{n+1}-x_{n}),\]因为 $x_{n+1}-x_{n}\ne 0$，所以$$x_{n+1}=\dfrac{x_{n}+4}{x_{n}+1},n\in\mathbb N^{*}.$$ 备注1 问题2 判断 $x_{n}$ 与 $2$ 的大小关系，并证明你的结论； 答案2 $x_{n}>2$ 解析2 由 $(1)$ 可得\[x_{n}-2=\dfrac{x_{n-1}+4}{x_{n-1}+1}-2=\dfrac{x_{n-1}-2}{x_{n-1}+1}.\]注意到 $x_{n}>0$，所以 $x_{n}-2$ 与 $x_{n-1}-2$ 异号．由于 $x_{1}=1<2$，所以 $x_{2}>2$．依次类推，当 $n$ 为奇数时 $x_{n}<2$，当 $n$ 为偶数时，$x_{n}>2$． 备注2 问题3 求证：$|x_{1}-2|+|x_{2}-2|+\cdots+|x_{n}-2|<2$． 答案3 略 解析3 由于$$x_{n}>0,x_{n+1}=\dfrac{x_{n}+4}{x_{n}+1}=1+\dfrac{3}{x_{n}+1},$$所以 $x_{n}\geqslant 1$（$n=1,2,3,\cdots$）．于是\[\left|x_{n+1}-2\right|=\left|\dfrac{x_{n}-2}{x_{n}+1}\right|=\dfrac{\left|x_{n}-2\right|}{x_{n}+1}\leqslant \dfrac{1}{2}\left|x_{n}-2\right|,\]则有\[\begin{split}\left|x_{n}-2\right|&\leqslant \dfrac{1}{2}\left|x_{n-1}-2\right|\\ &\leqslant \dfrac{1}{2^{2}}\left|x_{n-2}-2\right|\\ &\leqslant \cdots\\ &\leqslant \dfrac{1}{2^{n-1}}\left|x_{1}-2\right|=\dfrac{1}{2^{n-1}},\end{split}\]因此\[\begin{split}LHS&\leqslant 1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{2}+\cdots+\left(\dfrac{1}{2}\right)^{n-1}\\&=2-\left(\dfrac{1}{2}\right)^{n-1}<2,\end{split}\]命题得证． 备注3