7: Ta có: \(\sqrt{x^2-6x+1}=2x-1\)
\(\Leftrightarrow4x^2-4x+1-x^2+6x-1=0\)
\(\Leftrightarrow3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)