xy+2x-y=0
⇒x.(y+2)=y
Nếu y+2=0⇒y=-2⇒0x=-2 (loại)
CHia 2 vế cho y+2$\neq$0
⇒x=$\frac{y}{y+2}$
=>x=$\frac{(y+2)-2}{y+2}$
=>x=$1-\frac{2}{y+2}$
⇒y+2∈Ư(2)={±1;±2}
y+2=1⇒y=-1⇒x=-1
y+2=-1⇒y=-3⇒x=3
y+2=2⇒y=0⇒x=0
y+2=-2⇒y=-4⇒x=2
Vậy (x,y)∈{-1;2;);(3;-2);(0;0);(2;-4}