ĐK: $\left[\begin{array}{l}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\le-1\end{array}\right.$
Đặt: $u=\sqrt{2x^2+3x+1};v=\sqrt{1-3x},u,v\ge0$, phương trình trở thành:
$u+v=2\sqrt{u^2+v^2}$
$\Leftrightarrow u^2+2uv+v^2=2(u^2+v^2)$
$\Leftrightarrow u^2-2uv+v^2=0$
$\Leftrightarrow (u-v)^2=0$
$\Leftrightarrow u=v$
Khi đó ta có:
$\sqrt{2x^2+3x+1}=\sqrt{1-3x}$
$\Leftrightarrow 2x^2+3x+1=1-3x$
$\Leftrightarrow x^2+3x=0$
$\Leftrightarrow \left[\begin{array}{l}x=0\\x=-3\end{array}\right.$ (thoả mãn)