Question

16．不等式$\frac{x-1}{x+2}$＞0的解集是{x|x＞1或x＜-2}．

Answer 1

分析 不等式$\frac{x-1}{x+2}$＞0即為$\left\{\begin{array}{l}{x-1＞0}\\{x+2＞0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜0}\\{x+2＜0}\end{array}\right.$，由一次不等式的解法，即可得到解集．

解答 解：不等式$\frac{x-1}{x+2}$＞0即為
$\left\{\begin{array}{l}{x-1＞0}\\{x+2＞0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜0}\\{x+2＜0}\end{array}\right.$，
解得x＞1或x＜-2．
則解集為{x|x＞1或x＜-2}．
故答案為：{x|x＞1或x＜-2}．

點(diǎn)評 本題考查分式不等式的解法，可以運(yùn)用符號法則或化為整式不等式，注意等價變形，屬于基礎(chǔ)題．