Đáp án: $x<\dfrac{-\sqrt{37}-3}{2}\quad \mathrm{hoặc}\quad \:x>\dfrac{\sqrt{37}-3}{2}$
Giải thích các bước giải:
ĐKXĐ: $x\le -2$ hoặc $x\ge -1$
Ta có :
$\sqrt{x^2+3x+2}<x^2+3x-4$
$\to (x^2+3x+2)-\sqrt{x^2+3x+2}-6>0$
$\to (x^2+3x+2)-3\sqrt{x^2+3x+2}+2\sqrt{x^2+3x+2}-6>0$
$\to \sqrt{x^2+3x+2}(\sqrt{x^2+3x+2}-3)+2(\sqrt{x^2+3x+2}-3)>0$
$\to (\sqrt{x^2+3x+2}+2)(\sqrt{x^2+3x+2}-3)>0$
Vì $\sqrt{x^2+3x+2}+2>0$
$\to \sqrt{x^2+3x+2}-3>0$
$\to \sqrt{x^2+3x+2}>3$
$\to x^2+3x+2>9$
$\to x^2+3x-7>0$
$\to \left(x+\dfrac{3}{2}\right)^2-\dfrac{37}{4}>0$
$\to \left(x+\dfrac{3}{2}\right)^2>\dfrac{37}{4}$
$\to x<\dfrac{-\sqrt{37}-3}{2}\quad \mathrm{hoặc}\quad \:x>\dfrac{\sqrt{37}-3}{2}$
