\(Đkxđ:x\ge\frac{1}{3}\)
Ta có: \(2\left(x+2\right)\sqrt{3x-1}=3x^2-7x-3\)
\(\Leftrightarrow\left(x+2\right)^2+2\left(x+2\right)\sqrt{3x-1}+3x-1=4x^2\)
\(\Leftrightarrow\left(x+2+\sqrt{3x-1}\right)^2=4x^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2+\sqrt{3x-1}=2x\\x+2+\sqrt{3x-1}=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=x-2\\\sqrt{3x-1}=-3x-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\3x+1=\left(x-2\right)^2\end{matrix}\right.\left(Pt:\sqrt{3x-1}=-3x-2vn\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-7x+5=0\end{matrix}\right.\Leftrightarrow x=\frac{7+\sqrt{29}}{2}\)
Vậy nghiệm của pt là: \(x=\frac{7+\sqrt{29}}{2}\)
