xy-2x-2y=0
⇒x.(y-2)=2y
Nếu y-2=0⇒y=2⇒0x=4 (loại)
Chia 2 vế cho y-2≠0, ta được:
x=$\frac{2y}{y-2}$
⇒x=$\frac{2(y-2)+4}{y-2}$
⇒x=$2+\frac{4}{y-2}$
⇒y-2∈Ư(4)={±1;±2;±4}
y-2=1⇒y=3⇒x=6
y-2=-1⇒y=1⇒x=-2
y-2=2⇒y=4⇒x=4
y-2=-2⇒y=0⇒x=0
y-2=4⇒y=6⇒x=3
y-2=-4⇒y=-2⇒x=1
Vậy (x,y)∈{(6;3);(-2;1);(4;4);(0;0);(3;6);(1;-2)}