\(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{2a+9-\left[5a+17\right]-3a}{a+3}=\frac{2a+9-5a-17-3a}{a+3}=\frac{-6a-8}{a+3}\)
Để \(\frac{-6a-8}{a+3}\)nguyên thì
\(-6a-8⋮a+3\)
\(\Rightarrow-6\left[a+3\right]+10⋮a+3\)
\(\Rightarrow10⋮a+3\)
=> a + 3 \(\inƯ\left[10\right]\in\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow a\in\left\{-13;-8;-5;-4;-2;-1;2;7\right\}\)