Question

15．不等式|$\frac{{x}^{2}-1}{x+1}$|＞4的解集是{x|x＜-3或x＞5}．

Answer 1

分析 由已知条件把原不等式转化为$\left\{\begin{array}{l}{x-1＞4}\\{x+1≠0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜-4}\\{x+1≠0}\end{array}\right.$，由此能求出不等式|$\frac{{x}^{2}-1}{x+1}$|＞4的解集．

解答 解：∵|$\frac{{x}^{2}-1}{x+1}$|＞4，
∴$\frac{{x}^{2}-1}{x+1}$＞4或$\frac{{x}^{2}-1}{x+1}＜-4$，
∴$\left\{\begin{array}{l}{x-1＞4}\\{x+1≠0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜-4}\\{x+1≠0}\end{array}\right.$，
解得x＞5或x＜-3．
∴不等式|$\frac{{x}^{2}-1}{x+1}$|＞4的解集是{x|x＜-3或x＞5}．
故答案为：{x|x＜-3或x＞5}．

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