a : 3 dư 1 \(\Rightarrow a-1⋮3\)
b : 3 sư 2 \(\Rightarrow b-2⋮3\)
\(\Rightarrow\left(a-1\right)\left(b-2\right)=ab-\left(2a+b\right)+2⋮3\)
Ta có \(a-1⋮3\Rightarrow2a-2⋮3\)
\(\Rightarrow2a-2+b-2=2a+b-4=2a+b-1-3⋮3\Rightarrow2a+b-1⋮3\)
Từ \(ab-\left(2a+b\right)+2=ab-\left(2a+b-1\right)+1⋮3\)
Mà \(2a+b-1⋮3\Rightarrow ab+1⋮3\) => ab : 3 dư 2