Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$
PT $\Leftrightarrow 4x+1=\sqrt{3x+1}$
\(\Rightarrow \left\{\begin{matrix} 4x+1\geq 0\\ (4x+1)^2=3x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{4}\\ 16x^2+5x=0\end{matrix}\right.\)
\(\Rightarrow x=0\)

