Ta có: (a2 + 3a + 6)=(a\(^2\)+3a)+6=a.(a+3)+6
\(\left(a+3\right)⋮\left(a+3\right)\)
Mà \(a\inℤ\)\(\Rightarrow a.\left(a+3\right)⋮\left(a+3\right)\)
Để (a2 + 3a + 6) \(⋮\)(a + 3) thì \(6⋮\left(a+3\right)\)
\(\Rightarrow a+3\inƯ\left(6\right)\)
\(\Leftrightarrow a+3\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Leftrightarrow a\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\)
Vậy .....