Question

14．解不等式：|x-|2x+1||＞1．

Answer 1

分析 分情況進(jìn)行絕對(duì)值的化簡(jiǎn)，然后解不等式．

解答 解：不等式可化為：x-|2x+1|＞1或x-|2x+1|＜-1；
整理得：|2x+1|＜x-1或|2x+1|＞x+1，
由|2x+1|＜x-1可得$\left\{\begin{array}{l}{2x+1＞0}\\{2x+1＜x-1}\end{array}\right.$或$\left\{\begin{array}{l}{2x+1＜0}\\{-2x-1＜x+1}\end{array}\right.$，
無(wú)解；
由|2x+1|＞x+1可得$\left\{\begin{array}{l}{2x+1＞0}\\{2x+1＞x+1}\end{array}\right.$或$\left\{\begin{array}{l}{2x+1＜0}\\{-2x-1＞x+1}\end{array}\right.$，
解得x＞0或x＜-$\frac{2}{3}$，
故不等式的解集為：x＞0或x$＜-\frac{2}{3}$．

點(diǎn)評(píng) 本題考查了不等式的解法和絕對(duì)值的化簡(jiǎn)，解答本題的關(guān)鍵是根據(jù)x值的不同分情況進(jìn)行絕對(duì)值的化簡(jiǎn)．