Question

不等式x²-3|x|+2＞0的解集是(    )

A.{x|x＜1或x＞2}                              B.{x|-1＜x＜1或x＞2}

C.{x|x＜-2或-1＜x＜1或x＞2}            D.{x|x＜-1或x＞1}

Answer 1

C

解析:設t=|x|≥0,則原不等式為t²-3t+2＞0,解得t＜1或t＞2,即|x|＜1或|x|＞2,

∴-1＜x＜1或x＜-2或x＞2.