Để \(\dfrac{a+3}{a-1}\in N\) thì \(a+3⋮a-1\)
\(\Rightarrow\left(a-1\right)+4⋮a-1\)
\(\Rightarrow4⋮a-1\)
\(\Rightarrow a-1\in U\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)
Mà \(a\in N\) \(\Rightarrow a\in\left\{0;2;3;5\right\}\)
Vậy \(a\in\left\{0;2;3;5\right\}\) thì \(\dfrac{a+3}{a-1}\in N\)