2xy-2y+x=11
⇒2y(x-1)+(x-1)=10
⇒(x-1)(2y+1)=10
Th1: $\left \{ {{x-1=1} \atop {2y+1=10}} \right.$ =>$\left \{ {{x=2} \atop {y=9/2}} \right.$ (loại)
Th2: $\left \{ {{x-1=-1} \atop {2y+1=-10}} \right.$ =>$\left \{ {{x=0} \atop {y=-11/2}} \right.$ (loại)
Th3: $\left \{ {{x-1=10} \atop {2y+1=1}} \right.$ =>$\left \{ {{x=11} \atop {y=0}} \right.$ (tm)
Th4: $\left \{ {{x-1=-10} \atop {2y+1=-1}} \right.$ =>$\left \{ {{x=-9} \atop {y=-1}} \right.$ (tm)
Th5: $\left \{ {{x-1=5} \atop {2y+1=2}} \right.$ =>$\left \{ {{x=6} \atop {y=1/2}} \right.$ (loại)
Th6: $\left \{ {{x-1=-5} \atop {2y+1=-2}} \right.$ =>$\left \{ {{x=-4} \atop {y=-3/2}} \right.$ (loại)
Th7: $\left \{ {{x-1=2} \atop {2y+1=5}} \right.$ =>$\left \{ {{x=3} \atop {y=2}} \right.$ (tm)
Th8: $\left \{ {{x-1=-2} \atop {2y+1=-5}} \right.$ =>$\left \{ {{x=-1} \atop {y=-3}} \right.$ (tm)
Vậy (x,y)∈{(11;0);(-9;-1);(3;2);(-1;-3)}
