Giải thích các bước giải:
m)$|x^3-1|\ge 1-x$
$\to |x-1||x^2+x+1|\ge -(x-1)$
$\to |x-1|(x^2+x+1)\ge -(x-1)$
+) $x\ge 1\to |x-1|=x-1$
$\to (x-1)(x^2+x+1)\ge -(x-1)$
$\to (x-1)(x^2+x+2)\ge 0$
$\to x-1\ge 0\to x\ge 1$
+) $x<1\to |x-1|=-x+1$
$\to (1-x)(x^2+x+1)\ge (1-x)$
$\to (1-x)(x^2+x)\ge 0$
$\to x^2+x\ge 0$
$\to x(x+1)\ge 0$
$\to x\ge 0$ hoặc $x\le -1$
$\to 0\le x<1$ hoặc $x\le -1$
p) $|\dfrac{2x-1}{x-1}|>2$
$\to \dfrac{2x-1}{x-1}<-2\to \dfrac{2x-1}{x-1}+2<0\to \dfrac{4x-3}{x-1}<0\to \dfrac 34<x<1$
Hoặc $\dfrac{2x-1}{x-1}>2\to \dfrac{1}{x-1}>0\to x>1$
