Ta có: \(\dfrac{a^2+3a-7}{a+3}\)
\(\Leftrightarrow a-\dfrac{7}{a+3}\)
\(\Rightarrow a+3\inƯ_{\left(7\right)}=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}a=-10\\a=-4\\a=-2\\a=4\end{matrix}\right.\)
Vậy: \(a=-10;-4;-2;4\) thì \(\dfrac{a^2+3a-7}{a+3}\) có giá trị nguyên