ĐKXĐ: \(x\ne2;y\ne-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\2+\dfrac{2}{x-2}+1-\dfrac{1}{y+1}=\dfrac{14}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+\dfrac{2}{y+1}=\dfrac{11}{3}\\\dfrac{2}{x-2}-\dfrac{1}{y+1}=\dfrac{5}{3}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=u\\\dfrac{1}{y+1}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3u+2v=\dfrac{11}{3}\\2u-v=\dfrac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=1\\\dfrac{1}{y+1}=\dfrac{1}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
