ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\dfrac{x+2}{x-2}+\dfrac{1}{x}=\dfrac{-8}{2x-x^2}\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}+\dfrac{x-2}{x\left(x-2\right)}+\dfrac{8}{x\left(2-x\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x}{x\left(x-2\right)}+\dfrac{x-2}{x\left(x-2\right)}-\dfrac{8}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x+x-2-8}{x\left(x-2\right)}=0\\ \Rightarrow x^2+3x-10=0\\ \Leftrightarrow\left(x+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x+2}{x-2}+\dfrac{1}{x}-\dfrac{8}{x\left(x-2\right)}=0\) ( đkxđ : x khác 0 ; x khác 2
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}+\dfrac{1.\left(x-2\right)}{x\left(x-2\right)}-\dfrac{8}{x\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+2x+x-2-8=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
=> x = 2( ko tm)
x = -5 ( tm)
