We are asked to find the sum of the following terms
[tex]\frac{7x}{x^2-4}\; and\; \frac{2}{x+2}[/tex]Let us find their sum.
First of all, notice that the following denominator can be factored as
[tex]x^2-4=x^2-2^2=(x-2)(x+2)[/tex]So, the expression becomes
[tex]\begin{gathered} \frac{7x}{(x-2)(x+2)}+\frac{2}{x+2} \\ \end{gathered}[/tex]Add them by taking the LCM
[tex]\frac{7x}{(x-2)(x+2)}+\frac{2}{x+2}=\frac{7x+(x-2)\cdot2}{(x-2)(x+2)}[/tex]Simplify the expression
[tex]\frac{7x+(x-2)\cdot2}{(x-2)(x+2)}=\frac{7x+2x-4}{(x-2)(x+2)}=\frac{9x-4}{(x-2)(x+2)}[/tex]Finally, re-write the denominator in the expanded form as it was written before.
[tex]\frac{9x-4}{(x-2)(x+2)}=\frac{9x-4}{x^2-4}[/tex]Therefore, the simplified expression is
[tex]\frac{9x-4}{x^2-4}[/tex]