\(2xy-x-y=2\)
\(\Rightarrow4xy-2x-2y=4\)
\(\Rightarrow4xy-2x-2y+1=4+1\)
\(\Rightarrow\left(4xy-2x\right)-\left(2y-1\right)=5\)
\(\Rightarrow2x.\left(2y-1\right)-\left(2y-1\right)=5\)
\(\Rightarrow\left(2x-1\right).\left(2y-1\right)=5\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}2x-1\in Z\\2y-1\in Z\end{matrix}\right.\)
\(\Rightarrow2x-1\inƯC\left(5\right);2y-1\inƯC\left(5\right)\)
\(\Rightarrow2x-1\in\left\{\pm1;\pm5\right\};2y-1\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1=1\\2y-1=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1=5\\2y-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1=-1\\2y-1=-5\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1=-5\\2y-1=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x=2\\2y=6\end{matrix}\right.\\\left\{{}\begin{matrix}2x=6\\2y=2\end{matrix}\right.\\\left\{{}\begin{matrix}2x=0\\2y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}2x=-4\\2y=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=0\\y=-2\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\left(TM\right)\end{matrix}\right.\)
Vậy các cặp số nguyên \(\left(x;y\right)\) thỏa mãn đề bài là: \(\left(1;3\right),\left(3;1\right),\left(0;-2\right),\left(-2;0\right).\)
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