\(\left\{{}\begin{matrix}\dfrac{2a-9}{a+3}=\dfrac{2a+6-15}{a+3}=\dfrac{2a+6}{a+3}-\dfrac{15}{a+3}=2-\dfrac{15}{a+3}\\\dfrac{5a+17}{a+3}=\dfrac{5a+15+2}{a+3}=\dfrac{5a+15}{a+3}+\dfrac{2}{a+3}=5+\dfrac{2}{a+3}\\\dfrac{3a}{a+3}=\dfrac{3a+9-9}{a+3}=\dfrac{3a+9}{a+3}-\dfrac{9}{a+3}=3-\dfrac{9}{a+3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}15⋮a+3\\2⋮a+3\\9⋮a+3\end{matrix}\right.\) \(\Rightarrow a+3\inƯC\left(15;2;9\right)=\pm1\)
Vậy thỏa mãn khi \(a=\pm1\)