7: a, (x+2)(y-1)=3
TH1: $\left \{ {{x+2=1} \atop {y-1=3}} \right.$ ⇒$\left \{ {{x=-1} \atop {y=4}} \right.$
TH2: $\left \{ {{x+2=-1} \atop {y-1=-1}} \right.$ ⇒$\left \{ {{x=-2} \atop {y=0}} \right.$
TH3: $\left \{ {{x+2=3} \atop {y-1=1}} \right.$ ⇒$\left \{ {{x=1} \atop {y=2}} \right.$
TH4: $\left \{ {{x+2=-3} \atop {y-1=-1}} \right.$ ⇒$\left \{ {{x=-5} \atop {y=0}} \right.$
Vậy (x,y)∈{(-1;4);(-2;0);(1;2);(-5;0)}
b, (3-x)(xy+5)=-1
TH1: $\left \{ {{3-x=1} \atop {xy+5=-1}} \right.$ ⇒$\left \{ {{x=2} \atop {y=-3}} \right.$
TH2: $\left \{ {{3-x=-1} \atop {xy+5=1}} \right.$ ⇒$\left \{ {{x=4} \atop {y=-1}} \right.$
Vậy (x,y)∈{(2;-3);(4;-1)}
