Question

3．解不等式0＜x²-x-2≤4．

Answer 1

分析 不等式0＜x²-x-2≤4等價(jià)于$\left\{\begin{array}{l}{x^2}-x-2＞0\\{x^2}-x-2≤4\end{array}\right.$，求不等式組的解集即可．

解答 解：不等式0＜x²-x-2≤4等價(jià)于
$\left\{\begin{array}{l}{x^2}-x-2＞0\\{x^2}-x-2≤4\end{array}\right.$，
即$\left\{\begin{array}{l}{x^2}-x-2＞0\\{x^2}-x-6≤0\end{array}\right.$，
可化為$\left\{\begin{array}{l}（{x-2}）（{x+1}）＞0\\（{x-3}）（{x+2}）≤0\end{array}\right.$，
解得$\left\{\begin{array}{l}x＞2或x＜-1\\-2≤x≤3.\end{array}\right.$，

如圖所示；
所以原不等式的解集為{x|-2≤x＜-1或2＜x≤3}．

點(diǎn)評(píng) 本題考查了不等式組的解法與應(yīng)用問(wèn)題，是基礎(chǔ)題．