题目内容
12.解不等式:|x-2|+|2x-1|>x+5.分析 把要解的不等式等价转化为与之等价的三个不等式组,求出每个不等式组的解集,再取并集,即得所求.
解答 解:不等式:|x-2|+|2x-1|>x+5等价于$\left\{\begin{array}{l}{x<\frac{1}{2}}\\{2-x+1-2x>x+5}\end{array}\right.$①,或$\left\{\begin{array}{l}{\frac{1}{2}≤x<2}\\{2-x+2x-1>x+5}\end{array}\right.$②,或$\left\{\begin{array}{l}{x≥2}\\{x-2+2x-1>x+5}\end{array}\right.$ ③.
解①求得x<-$\frac{1}{2}$,解②求得x∈∅,解③求得x>4.
综上可得,原不等式的解集为{x|x<-$\frac{1}{2}$,或x>4}.
点评 本题主要考查绝对值不等式的解法,体现了转化、分类讨论的数学思想,属于基础题.
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