Đáp án: $ \:x<-1\quad \mathrm{hoặc}\quad \dfrac{3-\sqrt{41}}{8}\le \:x<0$
Giải thích các bước giải:
Ta có:
$\dfrac{4x-1}{x+1}-1\ge\dfrac{2-x}{x}$
$\to \dfrac{4x-1}{x+1}-1\ge \dfrac2x-1$
$\to \dfrac{4x-1}{x+1}\ge \dfrac2x$
$\to \dfrac{4x-1}{x+1}- \dfrac2x\ge 0$
$\to \dfrac{x(4x-1)-2(x+1)}{x(x+1)}\ge 0$
$\to \dfrac{4x^2-3x-2}{x(x+1)}\ge 0$
$\to \dfrac{4(x-\dfrac{3+\sqrt{41}}{8})(x-\dfrac{3-\sqrt{41}}{8})}{x(x+1)}\ge 0$
$\to \:x<-1\quad \mathrm{hoặc}\quad \dfrac{3-\sqrt{41}}{8}\le \:x<0$
