\(xy-5x+3y=7\\ \Rightarrow x\left(y-5\right)+3y-15=7-15\\ \Rightarrow x\left(y-5\right)+3\left(y-5\right)=-8\\ \Rightarrow\left(x+3\right)\left(y-5\right)=-8\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x+3,y-5\in Z\\x+3,y-5\inƯ\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\end{matrix}\right.\)
Ta có bảng:
| x+3 | -1 | -2 | -4 | -8 | 1 | 2 | 4 | 8 |
| y-5 | 8 | 4 | 2 | 1 | -8 | -4 | -2 | -1 |
| x | -4 | -5 | -7 | -11 | -2 | -1 | 1 | 5 |
| y | 13 | 9 | 7 | 6 | -3 | 1 | 3 | 4 |
\(\text{Vậy }\left(x,y\right)\in\left\{\left(-4;13\right);\left(-5;9\right);\left(-7;7\right);\left(-11;6\right);\left(-2;-3\right);\left(-1;1\right);\left(1;3\right);\left(5;4\right)\right\}\)
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