b)
[tex]\it \dfrac{2n-3}{2n+3}= \dfrac{2n+3-6}{2n+3}= \dfrac{2n+3}{2n+3}- \dfrac{6}{2n+3} =1- \dfrac{6}{2n+3}\ \in\mathbb{Z}\Rightarrow\\ \\ \\ \Rightarrow \dfrac{6}{2n+3}\in\mathbb{Z}\ \ \ \ (1)\\ \\ \\ 2n+3=impar\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow 2n+3\in\{-3,\ -1,\ 1,\ 3\}|_{-3} \Rightarrow 2n\in\{-6,\ -4,\ -2,\ 0\}|_{:2}\Rightarrow\\ \\ \\ \Rightarrow n\in\{-3,\ -2,\ -1,\ 0\}[/tex]
