Ta có: \(2x^2-4x+\sqrt{x^2-2x+1}=1\)
<=> \(2x^2-4x+\sqrt{\left(x-1\right)^2}=1\)
<=> \(2x^2-4x+x-1=1\)
<=> \(2x^2-3x-2=0\)
<=> \(\left(2x^2-4x\right)+\left(x-2\right)=0\)
<=> \(2x\left(x-2\right)+\left(x-2\right)=0\)
<=> \(\left(x-2\right)\left(2x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\) => x = 2