a) \(xy=x+y\Rightarrow xy-x-y=0\)
\(\Rightarrow xy-x-y+1=0+1=1\)
\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=1\)
\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)
\(\Rightarrow x-1;y-1\inƯ\left(1\right)=\left\{-1;1\right\}\)
x-1 | -1 | 1 |
x | 0 | 2 |
y-1 | -1 | 1 |
y | 0 | 2 |
Vậy (x;y)={(0;0);(2;2)}