Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$
PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$
$\Leftrightarrow \frac{x}{\sqrt{x+2}}=\frac{2x}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow x\left(\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}\right)=0$
Xét các TH:
TH1: $x=0$ (thỏa mãn)
TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow \sqrt{3x+1}+\sqrt{x+1}=2\sqrt{x+2}$
$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$
$\Leftrightarrow \sqrt{(3x+1)(x+1)}=3$
$\Rightarrow (3x+1)(x+1)=9$
$\Leftrightarrow 3x^2+4x-8=0$
$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$
Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$
Vậy............
