ĐKXĐ: ...
\(\Leftrightarrow3x^2+3x-1=\left(3x+1\right)\sqrt{x^2+2x-1}\)
\(\Leftrightarrow x^2+2x-1-\left(3x+1\right)\sqrt{x^2+2x-1}+2x^2+x=0\)
Đặt \(\sqrt{x^2+2x-1}=t\ge0\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+x\right)=\left(x+1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{3x+1-\left(x+1\right)}{2}=x\\t=\frac{3x+1+x+1}{2}=2x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=x\left(x\ge0\right)\\\sqrt{x^2+2x-1}=2x+1\left(x\ge-\frac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x^2\\x^2+2x-1=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow...\)
