Giải thích các bước giải:
a.Ta có $(x-1)^2\ge 0$
$\to (x-1)^2(x+1)(2-x)\ge 0$
$\to (x+1)(2-x)\ge 0$
$\to (x+1)(x-2)\le 0$
$\to -1\le x\le 2$
b.Ta có:
$\dfrac{1}{x^2-3x-4}\ge\dfrac1{1-x}$
$\to \dfrac{1}{x^2-3x-4}-\dfrac1{1-x}\ge0$
$\to \dfrac{1}{x^2-3x-4}+\dfrac1{x-1}\ge0$
$\to \dfrac{x^2-2x-5}{\left(x+1\right)\left(x-4\right)\left(x-1\right)}\ge \:0$
$\to \dfrac{\left(x-1-\sqrt{6}\right)\left(x-1+\sqrt{6}\right)}{\left(x+1\right)\left(x-4\right)\left(x-1\right)}\ge \:0$
$\to -\sqrt{6}+1\le \:x<-1\quad \mathrm{hoặc}\quad \:1<x\le \:1+\sqrt{6}\quad \mathrm{hoặc}\quad \:x>4$
