Ta có : \(x^2+8x-20=\left(x-2\right)\left(x+10\right)\)
\(\left|x-2\right|=x-2\Leftrightarrow x\ge0\)
\(\left|x-2\right|=-\left(x-2\right)\Leftrightarrow x\le0\)
Vì \(x\ge0\)suy ra : \(\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}=\frac{x}{x+10}\)
Vì \(x\le0\)suy ra : \(\frac{x\left[-\left(x-2\right)\right]}{\left(x-2\right)\left(x+10\right)}=\frac{-x}{x+10}\)