ta có: a-1/7 = -3/b+3
=> (a-1).(b+3)/7.(b+3)=-3.7/(b+3).7
=>(a-1).(b+3)/7b+21=-21/7b+21
=>(a-1).(b+3)=-21
=> -21 chia hết cho (b+3)
=>(b+3) ∈ Ư(-21)
=>(b+3)∈ {±1;±3;±7;±21}
ta có bang sau
b+3|a-1|b |a
1 |21 |-2 |22
3 |7 |0 |8
7 |3 |4 |4
21 |1 |18 |2
-1 |-21 |-4 |-20
-3 |-7 |-6 |-6
-7 |-3 |-10 |-2
-21 |-1 |-24 |0
vậy (a;b) ∈{22;-2);(8;0);(4;4);(2;18);(-20;-4);(-6;-6);(-2;-10);(-24;-1)}
