ĐKXĐ: x<>2; y<>1
\(\left\{{}\begin{matrix}\dfrac{1}{x-2}-\dfrac{1}{y-1}=2\\\dfrac{2}{x-2}+\dfrac{3}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-2}-\dfrac{2}{y-1}=4\\\dfrac{2}{x-2}+\dfrac{3}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}-\dfrac{2}{y-1}-\dfrac{2}{x-2}-\dfrac{3}{y-1}=4-1\\\dfrac{1}{x-2}-\dfrac{1}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{5}{y-1}=3\\\dfrac{1}{x-2}=\dfrac{1}{y-1}+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=-\dfrac{5}{3}\\\dfrac{1}{x-2}=1:\dfrac{-5}{3}+2=2-\dfrac{3}{5}=\dfrac{7}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{5}{3}+1=-\dfrac{2}{3}\\x-2=\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{3}\\x=\dfrac{5}{7}+2=\dfrac{19}{7}\end{matrix}\right.\left(nhận\right)\)
