Ta có:\(\sqrt{x^2-4x+4}=-x+4\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=-x+4\)
\(\Leftrightarrow\left|x-2\right|=-x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=-x+4\\x-2=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\0=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3\)
Ta có: \(\sqrt{x^2-4x+4}=-x+4\)
\(\Leftrightarrow\left|x-2\right|=4-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=4-x\left(x\ge2\right)\\x-2=x-4\left(x< 2\right)\left(loại\right)\end{matrix}\right.\Leftrightarrow x+x=4+2\)
\(\Leftrightarrow2x=6\)
hay x=3(nhận)