xy+12=x+y
⇒xy-x=y-12
⇒x(y-1)=y-12
Nếu y-1=0⇒y=1⇒0x=-11 (loại)
Chia 2 vế cho y-1$\neq$0
⇒x=$\frac{y-12}{y-1}$
=>x=$1-\frac{11}{y-1}$
⇒y-1∈Ư(11)={±1;±11}
y-1=1⇒y=2⇒x=-10
y-1=-1⇒y=0⇒x=12
y-1=11⇒y=12⇒x=0
y-1=-11⇒y=-10⇒x=2
Vậy (x,y)∈{(-10;2);(12;0);(0;12);(2;-10)}
