Đáp án: $x\in\{0,-2\}$
Giải thích các bước giải:
Đặt $x^2+2x=t, t\ge 0$
Ta có:
$\sqrt{3x^2+6x+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}$
$\to \sqrt{3(x^2+2x)+16}+\sqrt{x^2+2x}=2\sqrt{(x^2+2x)+4}$
$\to \sqrt{3t+16}+\sqrt{t}=2\sqrt{t+4}$
$\to (\sqrt{3t+16}+\sqrt{t})^2=(2\sqrt{t+4})^2$
$\to 3t+16+2\sqrt{3t+16}\cdot\sqrt{t}+t=4(t+4)$
$\to 4t+16+2\sqrt{3t+16}\cdot\sqrt{t}=4t+16$
$\to 2\sqrt{3t+16}\cdot\sqrt{t}=0$
$\to t=0$ vì $t\ge 0\to 3t+16>0$
$\to x^2+2x=0$
$\to x(x+2)=0$
$\to x\in\{0,-2\}$
