f: ĐKXĐ: x<>y và x<>-1
\(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{1}{x-y}=3\\\dfrac{2}{x+1}-\dfrac{3}{x-y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{3}{x-y}=9\\\dfrac{2}{x+1}-\dfrac{3}{x-y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}=10\\\dfrac{1}{x+1}+\dfrac{1}{x-y}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1=\dfrac{1}{2}\\\dfrac{1}{x-y}=3-\dfrac{1}{x+1}=3-1:\dfrac{1}{2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=x-1=-\dfrac{3}{2}\end{matrix}\right.\left(nhận\right)\)
