\(x+x\cdot y+y=9\)
\(x\cdot\left(1+y\right)+\left(1+y\right)-1=9\)
\(\left(1+y\right)\cdot\left(x+1\right)=9-1\)
\(\left(1+y\right)\cdot\left(x+1\right)=8\)
vì x;y thuộc Z
suy ra \(1+y;x+1\)thuộc Z
suy ra \(1+y;x+1\)thuộc \(Ư\left(8\right)\)
Ta có bảng:
x+1 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
1+y | 8 | -8 | 4 | -4 | 2 | -2 | 1 | -1 |
x | 0 | -2 | 1 | -3 | 3 | -5 | 7 | -9 |
y | 7 | -9 | 3 | -5 | 1 | -3 | 0 | -2 |
Vậy \(\left(x;y\right)\)thuộc\(\left\{\left(0;7\right);\left(-2;-9\right);\left(1;3\right);\left(-3;-5\right);\left(3;1\right);\left(-5;-3\right);\left(7;0\right);\left(-9;-2\right)\right\}\)