Đáp án: $x\le \dfrac{-\sqrt{271}+14}{5}\quad \mathrm{hoặc}\quad \:x\ge \dfrac{\sqrt{271}+14}{5}$
Giải thích các bước giải:
Ta có:
$8x(x-4)-15\ge 3x(x-2)+2x$
$\to 8x^2-32x-15\ge 3x^2-6x+2x$
$\to 8x^2-32x-15\ge 3x^2-4x$
$\to 5x^2-28x-15\ge 0$
$\to 5\left(x-\dfrac{14}{5}\right)^2-\dfrac{271}{5}\ge \:0$
$\to 5\left(x-\dfrac{14}{5}\right)^2\ge \dfrac{271}{5}$
$\to \left(x-\dfrac{14}{5}\right)^2\ge \dfrac{271}{25}$
$\to x-\dfrac{14}{5}\le \:-\sqrt{\dfrac{271}{25}}\quad \mathrm{hoặc}\quad \:x-\dfrac{14}{5}\ge \sqrt{\dfrac{271}{25}}$
$\to x\le \dfrac{-\sqrt{271}+14}{5}\quad \mathrm{hoặc}\quad \:x\ge \dfrac{\sqrt{271}+14}{5}$
