\(=x-4+\sqrt{\left(x-4\right)^2}\\ =\left[{}\begin{matrix}x-4+x-4\left(x\ge4\right)\\x-4+4-x\left(x< 4\right)\end{matrix}\right.\\ =\left[{}\begin{matrix}2x-8\left(x\ge4\right)\\0\left(x< 4\right)\end{matrix}\right.\)
\(x-4+\sqrt{16-8x+x^2}\)
= \(x-4+\sqrt{\left(4-x\right)^2}\)
= x - 4 + 4 - x
= x - x - 4 + 4
= 0
\(\displaystyle \begin{array}{{>{\displaystyle}l}} x-4+\sqrt{16-8x+x^{2}}\\ \rightarrow x-4-\sqrt{( 4-x)^{2}}\\ \rightarrow x-4-|4-x|\\ Với\ \left[ \begin{array}{l l} 4-x\geqslant 0\ \rightarrow \ x-4-4+x=2x-8 & \\ 4-x< 0\ \rightarrow \ x-4+4-x=0\ & \end{array} \right. \end{array}\)
