Answer :
[tex]\frac{a}{a^2-4}+\frac{2}{3a+6} = \frac{a}{(a-2)(a+2)} + \frac{2}{3(a+2)} = \\\\\frac{3a}{3(a-2)(a+2)}+\frac{2(a-2)}{3(a+2)(a-2)} = \\\\\frac{3a+2a-4}{3(a^2-4)} = \frac{5a-4}{3a^2-12}[/tex]
[tex]\frac{5}{a}+7 = \frac{5}{a}+\frac{7a}{a} = \frac{7a+5}{a}\\\\\\\frac{a}{ab}-\frac{3}{b} = \frac{1}{b} -\frac{3}{b} = \frac{1-3}{b} = -\frac{2}{b}\\\\\frac{6}{ab}+\frac{8}{a} = \frac{6}{ab}+\frac{8b}{ab} = \frac{6+8b}{ab}\\\\\frac{3a}{4b^2} - \frac{b}{6a} = \frac{9a^2}{12ab^2} - \frac{2b^3}{12ab^2} = \frac{9a^2-2b^3}{12ab^2}[/tex]
[tex]\frac{5}{a}+7 = \frac{5}{a}+\frac{7a}{a} = \frac{7a+5}{a}\\\\\\\frac{a}{ab}-\frac{3}{b} = \frac{1}{b} -\frac{3}{b} = \frac{1-3}{b} = -\frac{2}{b}\\\\\frac{6}{ab}+\frac{8}{a} = \frac{6}{ab}+\frac{8b}{ab} = \frac{6+8b}{ab}\\\\\frac{3a}{4b^2} - \frac{b}{6a} = \frac{9a^2}{12ab^2} - \frac{2b^3}{12ab^2} = \frac{9a^2-2b^3}{12ab^2}[/tex]
