\(\sqrt{2x^2+4x-1}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\2x^2+4x-1=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2+6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=-3+\sqrt{11}\\x=-3-\sqrt{11}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)
=> pt vô nghiệm