\(3xy+x-3y=5\\ \Rightarrow x\left(3y+1\right)-3y-1=5-1\\ \Rightarrow x\left(3y+1\right)-\left(3y-1\right)=4\\ \Rightarrow\left(x-1\right)\left(3y-1\right)=4\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-1,3y-1\in Z\\x-1,3y-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\end{matrix}\right.\)
Ta có bảng:
| x-1 | 1 | 2 | 4 | -1 | -2 | -4 |
| 3y-1 | 4 | 2 | 1 | -4 | -2 | -1 |
| x | 2 | 3 | 5 | 0 | -1 | -3 |
| y | \(\dfrac{5}{3}\left(loại\right)\) | 1 | \(\dfrac{2}{3}\left(loại\right)\) | -1 | \(-\dfrac{1}{3}\left(loại\right)\) | 0 |
Vậy \(\left(x,y\right)\in\left\{\left(3;1\right);\left(0;-1\right);\left(-3;0\right)\right\}\)
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