\(A=\dfrac{2a+9}{a+3}+\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}=\dfrac{7a+26-3a}{a+3}=\dfrac{4a+26}{a+3}\)
Để A là số nguyên thì \(4a+12+14⋮a+3\)
\(\Leftrightarrow a+3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)
hay \(a\in\left\{-2;-4;-1;-5;4;-10;11;-17\right\}\)