\(\dfrac{1}{x}+\dfrac{2}{x\left(x-2\right)}=\dfrac{x+2}{x-2}\left(x\ne0;x\ne2\right)\)
\(< =>\dfrac{x-2}{x\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}=\dfrac{x\left(x+2\right)}{x\left(x-2\right)}\)
suy ra
`x-2+2=x^2 +2x`
`<=>x^2 +2x-x+2-2=0`
`<=>x^2 +x=0`
`<=>x(x+1)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{2}{x\left(x-2\right)}=\dfrac{x+2}{x-2}\Leftrightarrow\dfrac{x-2}{x\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}=\dfrac{x\left(x+2\right)}{x\left(x-2\right)}\)
\(\Rightarrow x-2+2=x^2+2x\)
\(\Leftrightarrow-x^2-x=0\Leftrightarrow-x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\Leftrightarrow x=-1\end{matrix}\right.\)
Vay \(S=\left\{-1\right\}\)