\(\dfrac{2n}{n+3}+\dfrac{2}{n+3}=\dfrac{2n+2}{n+3}\in N\left(n\ne-3\right)\)
\(\Rightarrow2n+2⋮n+3\)
\(\Rightarrow2n+2-2n-6⋮n+3\)
\(\Rightarrow-4⋮n+3\)
\(\Rightarrow n+3\in U\left(4\right)=\left\{1;2;4\right\}\)
\(\Rightarrow n\in\left\{-2;-1;1\right\}\)
Vậy có 3 số nguyên thỏa mãn đề bài