\(\left\{{}\begin{matrix}\dfrac{2}{x-y}+\dfrac{3}{2x+y-1}=4\\\dfrac{4}{x-y}-\dfrac{3}{2x+y-1}=2\end{matrix}\right.\) ĐK : \(x-y\ne0;2x+y-1\ne0\)
Đặt : \(\dfrac{2}{x-y}=a\) ; \(\dfrac{4}{x-y}=b\) (*)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\2a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\end{matrix}\right.\)
Thay a = 2 , b = 2 vào (*) , Ta có :
\(\left\{{}\begin{matrix}\dfrac{2}{x-y}=2\\\dfrac{3}{2x+y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2.\left(x-y\right)=2\\2.\left(2x+y-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=2\\4x+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{6}\\y=\dfrac{1}{6}\end{matrix}\right.\)
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