\(xy-y+3x-2=0\)
\(\Leftrightarrow y\left(x-1\right)+3x-3+1=0\)
\(\Leftrightarrow y\left(x-1\right)+3\left(x-1\right)+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y+3\right)=-1\)
Do \(x;y\in Z\)
\(\Rightarrow x-1;y+3\in Z\)
Mà \(x-1;y+3\inƯ\left(-1\right)\)
\(\Rightarrow x-1;y+3\in\left\{1;-1\right\}\)
Ta có bảng sau :
| \(x-1\) | \(1\) | \(-1\) |
| \(y+3\) | \(-1\) | \(1\) |
| \(x\) | \(2\) | \(0\) |
| \(y\) | \(-4\) | \(-2\) |
Vậy \(\left(x,y\right)\in\left\{\left(2,-4\right);\left(0,-2\right)\right\}\)
:D
Đúng 0
Bình luận (0)
