\(xy-2x-y=-6\\ \Rightarrow x\left(y-2\right)-y+2=-6+2\\ \Rightarrow x\left(y-2\right)-\left(y-2\right)=-4\\ \Rightarrow\left(x-1\right)\left(y-2\right)=-4\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-1,y-2\in Z\\x-1,y-2\inƯ\left(-4\right)\end{matrix}\right.\)
Ta có bảng:
| x-1 | -1 | -2 | -4 | 1 | 2 | 4 |
| y-2 | 4 | 2 | 1 | -4 | -2 | -1 |
| x | 0 | -1 | -3 | 2 | 3 | 5 |
| y | 6 | 4 | 3 | -2 | 0 | 1 |
Vậy \(\left(x,y\right)\in\left\{\left(0;6\right);\left(-1;4\right);\left(-3;3\right);\left(2;-2\right);\left(3;0\right);\left(5;1\right)\right\}\)
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