ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-\frac{1}{3}\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\frac{9x^2\left(1+\sqrt{1+3x}\right)^2}{\left(1-\sqrt{1+3x}\right)^2\left(1+\sqrt{1+3x}\right)^2}=3x+1\)
\(\Leftrightarrow\frac{9x^2\left(1+\sqrt{1+3x}\right)^2}{9x^2}=3x+1\)
\(\Leftrightarrow\left(1+\sqrt{1+3x}\right)^2=3x+1\)
Đặt \(\sqrt{3x+1}=t\ge0\)
\(\left(t+1\right)^2=t^2\Leftrightarrow2t+1=0\Rightarrow t=-\frac{1}{2}< 0\left(l\right)\)
Vậy pt đã cho vô nghiệm