Question

$frac{3a^{2}+6a}{a^{2}-9}- frac{2a}{a-3}}{}$

 

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Answer 1

$frac{3a^2+6a}{a^2-9}-frac{2a}{a-3}=frac{3a(a+2)}{(a-3)(a+3)}-frac{2a}{a-3}=frac{3a(a+2)-2a(a+3)}{(a-3)(a+3)}=$

$=frac{3a^2+6a-4a^2-6a}{(a-3)(a+3)}=frac{-a^2}{(a-3)(a+3)}=frac{-a^2}{a^2-9}$