a, (x-3)(y-5)=7
TH1: $\left \{ {{x-3=1} \atop {y-5=7}} \right.$ =>$\left \{ {{x=4} \atop {y=12}} \right.$
TH2: $\left \{ {{x-3=7} \atop {y-5=1}} \right.$ =>$\left \{ {{x=10} \atop {y=6}} \right.$
TH3: $\left \{ {{x-3=-1} \atop {y-5=-7}} \right.$ =>$\left \{ {{x=2} \atop {y=-2}} \right.$
TH4: $\left \{ {{x-3=-7} \atop {y-5=-1}} \right.$ =>$\left \{ {{x=-4} \atop {y=4}} \right.$
Vậy (x,y)∈{(4;12);(10;6);(2;-2);(-4;4)}
b, (2x+1)(y-1)=3
TH1: $\left \{ {{2x+1=1} \atop {y-1=3}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH2: $\left \{ {{2x+1=-1} \atop {y-1=-3}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH3: $\left \{ {{2x+1=3} \atop {y-1=1}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH4: $\left \{ {{2x+1=-3} \atop {y-1=-1}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
c, (x-3)(2-y)=5
TH1: $\left \{ {{x-3=1} \atop {2-y=5}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH2: $\left \{ {{x-3=-1} \atop {2-y=-5}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH3: $\left \{ {{x-3=5} \atop {2-y=1}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
TH4: $\left \{ {{x-3=-5} \atop {2-y=-1}} \right.$ =>$\left \{ {{x=...} \atop {y=...}} \right.$
Phần b và c tự tìm (x,y) nha bạn!!! Làm tương tự như phần a nhé!
