\(2xy-4x-y=3\left(ĐK:x;y\in Z\right)\\ 2xy-4x-y+2=3+2\\ 2x\cdot\left(y-2\right)-\left(y-2\right)=5\\ \left(2x-1\right)\left(y-2\right)=5=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\\ \left\{{}\begin{matrix}2x-1=1=>x=1\\y-2=5=>y=7\end{matrix}\right.\\ \left\{{}\begin{matrix}2x-1=5=>x=3\\y-2=1=>y=3\end{matrix}\right.\\ \left\{{}\begin{matrix}2x-1=-1=>x=0\\y-2=-5=>y=-3\end{matrix}\right.\\ \left\{{}\begin{matrix}2x-1=-5=>x=-2\\y-2=-1=>y=1\end{matrix}\right.\\ \text{vậy }\left(x;y\right)\in\left\{\left(3;3\right);\left(0;-3\right);\left(-2;1\right);\left(1;7\right)\right\}\)
Đề thiếu rồi em, phải có điều kiện x;y nguyên mới tính được
