\(Đặt:\left\{{}\begin{matrix}a=\dfrac{1}{x-1}\left(x\ne1\right)\\b=\dfrac{1}{y+2}\left(y\ne-2\right)\end{matrix}\right.\\ Có:\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{1}{y+2}=1\\\dfrac{6}{x-1}-\dfrac{5}{y+2}=\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a+b=1\\6a-5b=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6a+2b=2\\6a-5b=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7b=\dfrac{7}{4}\\3a+b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{4}:7=\dfrac{1}{4}\\a=\dfrac{1-\dfrac{1}{4}}{3}=\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{x-1}=\dfrac{1}{4}\\b=\dfrac{1}{y+2}=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\left(TM\right)\\y=2\left(TM\right)\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(5;2\right)\)
