xy+x+y+1=11
\(\Rightarrow\)x(y+1)+(y+1)=11
\(\Rightarrow\)(x+1)(y+1)=11
\(\Rightarrow\)(x+1)(y+1)=1.11=-1.-11=11.1=-11.-1
do x;y\(\in\)Z nên ta có bảng sau:
| x+1 | 1 | 11 | -1 | -11 |
| y+1 | 11 | 1 | -11 | -1 |
| x | 0 | 10 | -2 | -12 |
| y | 10 | 0 | -12 | -2 |
vậy x;y\(\in\){(0;10);(10;0);(-2;-12);(-12;-2)}
(x-3).(y+2)=-5
\(\Rightarrow\)(x-3).(y+2)=-1.5=-5.1
do x;y\(\in\)Z nen ta co bang sau:
| x-3 | -1 | -5 |
| y+2 | 5 | 1 |
| x | 2 | -2 |
| y | 3 | -1 |
vay:(x;y)\(\in\){(2;3);(-2;-1)}