\(2xy-5x+7y=12\)

\(\Leftrightarrow y\left(2x+7\right)-5x=12\)

\(\Leftrightarrow y\left(2x+7\right)=12+5x\)\(\Leftrightarrow y=\frac{12+5x}{2x+7}\left(1\right)\)

Để y nguyên thì \(\frac{12+5x}{2x+7}\in Z\Rightarrow12+5x⋮2x+7\)

Ta thấy: \(2\left(12+5x\right)⋮2x+7\Rightarrow24+10x⋮2x+7\)

Lại có: \(5\left(2x+7\right)⋮2x+7\Rightarrow10x+35⋮2x+7\)

Do đó: \(10x+35-\left(24+10x\right)⋮2x+7\)\(\Rightarrow11⋮2x+7\)

=> \(2x+7\inƯ\left(11\right)\). Mà \(x\in Z\Rightarrow2x+7\in Z\Rightarrow2x+7\in\left\{1;11;-1;-11\right\}\)

\(\Rightarrow2x\in\left\{-6;4;-8;-18\right\}\)\(\Rightarrow x\in\left\{-3;2;-4;-9\right\}\)

 Thay vào (1); ta được: \(y\in\left\{-2;2;-8;3\right\}\)

Vậy các cặp nghiệm nguyên của phương trình là:

\(\left(x;y\right)\in\left\{\left(-3;-2\right);\left(2;2\right);\left(-4;-8\right);\left(-9;3\right)\right\}.\)

\(\Rightarrow2x-4xy+2y=0\\ \Rightarrow2x\left(1-2y\right)+2y-1=-1\\ \Rightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\\ \Rightarrow\left(2x-1\right)\left(2y-1\right)=1=1.1=\left(-1\right)\left(-1\right)\)

Với \(\left\{{}\begin{matrix}2x-1=1\\2y-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\rightarrow\left(1;1\right)\)

Với \(\left\{{}\begin{matrix}2x-1=-1\\2y-1=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\rightarrow\left(0;0\right)\)

Vậy các cặp \(\left(x;y\right)\) cần tìm là \(\left(1;1\right);\left(0;0\right)\)