ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-2\\y\ne2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{x+2}+\dfrac{3}{2-y}=-1\\\dfrac{1}{2x+4}-\dfrac{1}{y-2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+2}-\dfrac{3}{y-2}=-1\\\dfrac{1}{2\left(x+2\right)}-\dfrac{1}{y-2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+2}-\dfrac{3}{y-2}=-1\\\dfrac{3}{2\left(x+2\right)}-\dfrac{3}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}\cdot\dfrac{1}{x+2}=-4\\\dfrac{3}{x+2}-\dfrac{3}{y-2}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x+2}=-4:\dfrac{3}{2}=-\dfrac{8}{3}\\\dfrac{1}{x+2}-\dfrac{1}{y-2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=-\dfrac{3}{8}\\\dfrac{1}{y-2}=\dfrac{1}{x+2}+\dfrac{1}{3}=\dfrac{-8}{3}+\dfrac{1}{3}=-\dfrac{7}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{3}{8}-2=-\dfrac{19}{8}\left(nhận\right)\\y-2=-\dfrac{3}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{19}{8}\left(nhận\right)\\y=-\dfrac{3}{7}+2=\dfrac{11}{7}\left(nhận\right)\end{matrix}\right.\)
