Add $ \dfrac{1}{n+2} $ and $ \dfrac{1}{n-2} $ to get $ \dfrac{ \color{purple}{ 2n } }{ n^2-4 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ n-2 }$ and the second by $\color{blue}{ n+2 }$.

$$ \begin{aligned} \frac{1}{n+2} + \frac{1}{n-2} & = \frac{ 1 \cdot \color{blue}{ \left( n-2 \right) }}{ \left( n+2 \right) \cdot \color{blue}{ \left( n-2 \right) }} + \frac{ 1 \cdot \color{blue}{ \left( n+2 \right) }}{ \left( n-2 \right) \cdot \color{blue}{ \left( n+2 \right) }} = \\[1ex] &=\frac{ \color{purple}{ n-2 } }{ n^2 -\cancel{2n}+ \cancel{2n}-4 } + \frac{ \color{purple}{ n+2 } }{ n^2 -\cancel{2n}+ \cancel{2n}-4 } = \\[1ex] &=\frac{ \color{purple}{ 2n } }{ n^2-4 } \end{aligned} $$

Add $ \dfrac{2n}{n^2-4} $ and $ \dfrac{3}{n^2-4} $ to get $ \dfrac{ 2n + 3 }{ \color{blue}{ n^2-4 }}$.

To add expressions with the same denominators, we add the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{2n}{n^2-4} + \frac{3}{n^2-4} & = \frac{2n}{\color{blue}{n^2-4}} + \frac{3}{\color{blue}{n^2-4}} =\frac{ 2n + 3 }{ \color{blue}{ n^2-4 }} \end{aligned} $$