Question

16-a^2/a^2-3a:a^2+4a/a^2-9

Answer 1

$frac{16-a^2}{a^2-3a} : frac{a^2+4a}{a^2-9} =frac{(4-a)(4+a)}{a (a-3)} : frac{a (a+4)}{(a-3)(a+3)} =frac{(4-a)(4+a)}{a (a-3)} * frac{(a-3)(a+3)}{a (a+4)} = \ = frac{(4-a)(a+3)}{a^2} = frac{4a+12-a^2-3a}{a^2} = frac{-a^2+a+12}{a^2}$