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)