tu de bai ta co \(\hept{\begin{cases}7+8+a+b+2+6⋮3.\left(1\right)\\a=4+b.\left(2\right)\end{cases}}\)
the (2) vao (1) duoc \(\left(23+4+2b\right)⋮3\) <=> \(\left(27+2b\right)⋮3\)
=> \(2b⋮3\) (do 27 chia het cho 3)
ma 2 ko chia het cho 3 => \(b⋮3\)
=> \(b\in\left\{0,3,6,9\right\}=>a\in\left\{4,7,10,13\right\}\Rightarrow\left(a;b\right)=\left(4;0\right),\left(7;3\right)\)
vay cac so a,b can tim la (4,0) , (7,3)