Giải:
\(x^2+xy-3x-2y-5=0\)
\(y.\left(x-2\right)=5+3x-x^2\)
\(y=\dfrac{-x^2+3x+5}{x-2}\)
\(\Rightarrow x-2\inƯ\left(-x^2+3x+5\right)\)
\(\Rightarrow x-2\inƯ\left(\left(x-2\right).\left(-x+1\right)+7\right)\)
\(\Rightarrow x-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Ta có bảng giá trị:
x-2=-7 thì y=-1
x=-5
x-2=-1 thì y=-7
x=1
x-2=1 thì y=7
x=3
x-2=7 thì y=1
x=9
Vậy (x;y)=(-5;-1);(1;-7);(3;7);(9;1)