Question

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

Answer 1

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