Question

11．不等式|x|＞$\frac{2}{x-1}$的解集为{x|x＜0或0＜x＜1 或x＞2}．

Answer 1

分析 不等式即$\left\{\begin{array}{l}{x≥0}\\{x＞\frac{2}{x-1}}\end{array}\right.$①，或 $\left\{\begin{array}{l}{x＜0}\\{-x＞\frac{2}{x-1}}\end{array}\right.$②，分别求得①、②的解集，再取并集，即得所求．

解答 解：不等式|x|＞$\frac{2}{x-1}$，即$\left\{\begin{array}{l}{x≥0}\\{x＞\frac{2}{x-1}}\end{array}\right.$①，或 $\left\{\begin{array}{l}{x＜0}\\{-x＞\frac{2}{x-1}}\end{array}\right.$②．
由①可得$\left\{\begin{array}{l}{x≥0}\\{\frac{（x+1）（x-2）}{x-1}＞0}\end{array}\right.$，求得0＜x＜1 或x＞2．
由②可得 $\left\{\begin{array}{l}{x＜0}\\{\frac{{（x-\frac{1}{2}）}^{2}+\frac{7}{4}}{x-1}＜0}\end{array}\right.$，求得 x＜0．
综上可得，原不等式的解集为{x|x＜0或0＜x＜1 或x＞2}，
故答案为：{x|x＜0或0＜x＜1 或x＞2}．

点评 本题主要考查绝对值不等式的解法，体现了转化、分类讨论的数学思想，属于中档题．