Question

11．不等式组$\left\{\begin{array}{l}{|x|=x}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$的解集是[0，$\sqrt{6}$]．

Answer 1

分析 把所给的绝对值不等式分类讨论，等价转化为$\left\{\begin{array}{l}{0≤x＜2}\\{\frac{3-x}{x+3}＞\frac{2-x}{x+2}}\end{array}\right.$ ①，或 $\left\{\begin{array}{l}{x≥2}\\{\frac{3-x}{x+3}＞\frac{x-2}{x+2}}\end{array}\right.$ ②．在分别求得①和②的解集，再取并集，即得所求．

解答 解：不等式组$\left\{\begin{array}{l}{|x|=x}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$，即$\left\{\begin{array}{l}{x≥0}\\{\frac{3-x}{x+3}＞\frac{|x-2|}{x+2}}\end{array}\right.$，即 $\left\{\begin{array}{l}{0≤x＜2}\\{\frac{3-x}{x+3}＞\frac{2-x}{x+2}}\end{array}\right.$ ①，或 $\left\{\begin{array}{l}{x≥2}\\{\frac{3-x}{x+3}＞\frac{x-2}{x+2}}\end{array}\right.$ ②．
解①可得0≤x＜2，解②可得 2≤x＜$\sqrt{6}$，
综上可得，不等式的解集为[0，$\sqrt{6}$]，
故答案为：[0，$\sqrt{6}$]．

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