a.(b-2)=3
TH1: $\left \{ {{a=1} \atop {b-2=3}} \right.$ =>$\left \{ {{a=1} \atop {b=5}} \right.$
TH2: $\left \{ {{a=-1} \atop {b-2=-3}} \right.$ =>$\left \{ {{a=11} \atop {b=-1}} \right.$
TH3: $\left \{ {{a=3} \atop {b-2=1}} \right.$ =>$\left \{ {{a=3} \atop {b=3}} \right.$
TH4: $\left \{ {{a=-3} \atop {b-2=-1}} \right.$ =>$\left \{ {{a=-3} \atop {b=1}} \right.$
Vậy (a,b)∈{1;5);(11;-1);(3;3);(-3;1)}
Bài 2:
Ta có: 2n+1$\vdots$n-3
⇒2(n-3)+7$\vdots$
⇒n-3∈Ư(7)={±1;±7}
n-3=1⇒n=4
n-3=-1⇒n=2
n-3=7⇒n=10
n-3=-7⇒n=-4
Vậy n∈{4;2;10;-4}