Đk $x \le \frac 13$
Đặt $x^2+1=a,1-3x=b$$pt\Leftrightarrow \sqrt{2a-b}+\sqrt b=2\sqrt a$
$\Leftrightarrow \sqrt{2a-b}-\sqrt a=\sqrt a-\sqrt b$
$\Leftrightarrow \frac{a-b}{\sqrt{2a-b}+\sqrt a}=\frac{a-b}{\sqrt a+\sqrt b}$
$\Leftrightarrow \left[ \begin{array}{l} a=b\\ \sqrt{2a-b}=\sqrt b \end{array} \right.\Leftrightarrow a=b$
$\Leftrightarrow x^2+1=1-3x\Leftrightarrow \left[ \begin{array}{l} x=0\\ x=-3 \end{array} \right.$