b, Ta có:
\(xy+2x-y=5\)
\(\Rightarrow\) \(xy+2x-y-2=5-2\)
\(\Rightarrow\left(xy-y\right)+\left(2x-2\right)=3\)
\(\Rightarrow y\left(x-1\right)+2\left(x-1\right)=3\)
\(\Rightarrow\left(y+2\right)\left(x-1\right)=3\)
\(\Rightarrow\left\{\left(y+2\right)\left(x-1\right)\right\}\inƯ_{\left(3\right)}\)
\(\Rightarrow\left\{\left(y+2\right)\left(x-1\right)\right\}\in\left\{\left(3;1\right)\left(1;3\right)\left(-1;-3\right)\left(-3;-1\right)\right\}\)
Ta có bảng sau:
| \(y+2\) | \(3\) | \(1\) | \(-3\) | \(-1\) |
| \(y\) | \(1\) | \(-1\) | \(-5\) | \(-3\) |
| \(x-1\) | \(1\) | \(3\) | \(-1\) | \(-3\) |
| \(x\) | \(2\) | \(4\) | \(0\) | \(-2\) |
- Các số trên thỏa mãn điều kiện: \(x;y\in Z\)
\(\Rightarrow\left\{\left(x;y\right)\right\}\in\left\{\left(2;1\right)\left(4;-1\right)\left(0;-5\right)\left(-2;-3\right)\right\}\)
Vậy \(\left\{\left(x;y\right)\right\}\in\left\{\left(2;1\right)\left(4;-1\right)\left(0;-5\right)\left(-2;-3\right)\right\}\)
Phần a tớ chưa nghĩ ra 
