\(x-2y-\sqrt{x^2-4xy+4y^2}\left(1\right)=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
TH1: \(x\ge2y\)
\(\left(1\right)=x-2y-x+2y=0\)
TH2: \(x< 2y\)
\(\left(1\right)=x-2y+x-2y=2x-4y\)
= x - 2y - \(\sqrt{\left(x-2y\right)^2}\)
= x - 2y - /x-2y/
= x - 2y - x + 2y
= 0
\(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\left|x-2y\right|\)
\(=\left[{}\begin{matrix}x-2y-x+2y=0\left(x\ge2y\right)\\x-2y+x-2y=2x-4y\left(x< 2y\right)\end{matrix}\right.\)