Giải thích các bước giải:
c.Ta có:
$(x+1)(2x-2)-3\ge -5x+(2x+1)(3-x)$
$\to 2x^2-2-3\ge -5x-2x^2+5x+3$
$\to 2x^2-5\ge -2x^2+3$
$\to 4x^2\ge 8$
$\to x^2\ge 2$
$\to x\ge \sqrt{2}$ hoặc $x\le -\sqrt{2}$
d.Ta có:
$\dfrac{x+17}{5}+2<\dfrac{3x-7}{4}$
$\to \dfrac{x+17}{5}\cdot \:20+2\cdot \:20<\dfrac{3x-7}{4}\cdot \:20$
$\to 4\left(x+17\right)+40<5\left(3x-7\right)$
$\to 4x+68+40<15x-35$
$\to 4x+108<15x-35$
$\to 11x>143$
$\to x>13$
