Question

7．解不等式组：$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$．

Answer 1

分析 由已知得$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞\frac{2-x}{2+x}}\end{array}\right.$或$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞\frac{x-2}{2+x}}\end{array}\right.$，由此能求出原不等式的解集．

解答 解：∵$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞\frac{2-x}{2+x}}\end{array}\right.$或$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞\frac{x-2}{2+x}}\end{array}\right.$，
解得x＞0或0＜x＜$\sqrt{6}$，
∴原不等式的解集为（0，+∞）．

点评 本题考查不等式组的解集的求法，是基础题，解题时要注意等价转化思想的合理运用．