ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-2\end{matrix}\right.\)
Ta có : \(\frac{2}{x-1}=1+\frac{2x}{x+2}\)
=> \(\frac{2}{x-1}=\frac{2x+x+2}{x+2}\)
=> \(\left(3x+2\right)\left(x-1\right)=2\left(x+2\right)\)
=> \(3x^2+2x-3x-2=2x+4\)
=> \(3x^2+2x-3x-2-2x-4=0\)
=> \(3x^2-3x-6=0\)
=> \(3x^2+3x-6x-6=0\)
=> \(3x\left(x+1\right)-6\left(x+1\right)=0\)
=> \(\left(3x-6\right)\left(x+1\right)=0\)
=> \(\left[{}\begin{matrix}3x-6=0\\x+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{2,-1\right\}\)
