Simplify $ \dfrac{x+2}{x^2-4} $ to $ \dfrac{1}{x-2} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$.

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

Add $ \dfrac{1}{x-2} $ and $ \dfrac{x-2}{x+2} $ to get $ \dfrac{ \color{purple}{ x^2-3x+6 } }{ x^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}{ x+2 }$ and the second by $\color{blue}{ x-2 }$.

$$ \begin{aligned} \frac{1}{x-2} + \frac{x-2}{x+2} & = \frac{ 1 \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x-2 \right) \cdot \color{blue}{ \left( x+2 \right) }} + \frac{ \left( x-2 \right) \cdot \color{blue}{ \left( x-2 \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( x-2 \right) }} = \\[1ex] &=\frac{ \color{purple}{ x+2 } }{ x^2+ \cancel{2x} -\cancel{2x}-4 } + \frac{ \color{purple}{ x^2-2x-2x+4 } }{ x^2+ \cancel{2x} -\cancel{2x}-4 } = \\[1ex] &=\frac{ \color{purple}{ x^2-3x+6 } }{ x^2-4 } \end{aligned} $$