Question

17．解不等式：$\frac{2x+1}{x-2}$＞0．

Answer 1

分析 将原不等式转化为$\left\{\begin{array}{l}{2x+1＞0}\\{x-2＞0}\end{array}\right.$或$\left\{\begin{array}{l}{2x+1＜0}\\{x-2＜0}\end{array}\right.$，由一次不等式的解法，即可得到所求解集．

解答 解：不等式：$\frac{2x+1}{x-2}$＞0即为
$\left\{\begin{array}{l}{2x+1＞0}\\{x-2＞0}\end{array}\right.$或$\left\{\begin{array}{l}{2x+1＜0}\\{x-2＜0}\end{array}\right.$，
即有$\left\{\begin{array}{l}{x＞-\frac{1}{2}}\\{x＞2}\end{array}\right.$或$\left\{\begin{array}{l}{x＜-\frac{1}{2}}\\{x＜2}\end{array}\right.$，
解得x＞2或x＜-$\frac{1}{2}$，
则解集为（-∞，-$\frac{1}{2}$）∪（2，+∞）．

点评 本题考查分式不等式的解法，考查符号法的运用，注意转化思想的运用，属于基础题．