\(xy-x-y=2\)
\(\Rightarrow xy-x-y+1=3\)
\(\Rightarrow x\left(y-1\right)-1\left(y-1\right)=3\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=3\)
\(\Rightarrow x-1;y-1\inƯ\left(3\right)\)
\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\Rightarrow x=2\\y-1=3\Rightarrow y=4\\x-1=-1\Rightarrow x=0\\y-1=-3\Rightarrow y=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=3\Rightarrow x=4\\y-1=1\Rightarrow y=2\\x-1=-3\Rightarrow x=-2\\y-1=-1\Rightarrow y=0\end{matrix}\right.\)
Ta có : xy - x - y = 2
=> x( y - 1) - y + 1 = 3
=> x(y - 1) - (y - 1) = 3
=> (x - 1) ( y - 1) = 3
Ta có bảng :
| x - 1 | 1 | 3 | -1 | -3 |
| y - 1 | 3 | 1 | -3 | -1 |
| x | 2 | 4 | 0 | -2 |
| y | 4 | 2 | -2 | 0 |
Vậy có 4 cặp (x;y) ....
