Ta có: $\frac{a^2+3ab}{a^2-9b^2}+\frac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\frac{a(a+3b)}{(a-3b)(a+3b)}-\frac{(a-3b)(2a+b)}{(a-3b)^2}$
$=\frac{a}{a-3b}-\frac{2a+b}{a-3b} =\frac{-a^2+2ab+3b^2}{a^2-9b^2}=-\frac{(a+b)(a-3b)}{(a-3b)(a+3b) }=-\frac{a+b}{a+3b}$
$\frac{a^2+an+ab+bn}{3bn-a^2-an+3ab}=-\frac{(a+b)(a+n)}{(a+n)(a-3b)}=-\frac{a+b}{a+3b} $
$\Rightarrow \frac{a^2+3ab}{a^2-9b^2}+\frac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\frac{a^2+an+ab+bn}{3bn-a^2-an+3ab}.$