Номер 2 (б,г,ж) . Срочно пожалуйста:)

Б)
$\frac{y^2+7y+12}{x^2-x-6}: \frac{y^2-16}{x^2-9}= \frac{(y^2+4y)+(3y+12)}{(x^2-3x)+(2x-6)}* \frac{(x-3)(x+3)}{(y-4)(y+4)}= \\ \\ = \frac{y(y+4)+3(y+4)}{x(x-3)+2(x-3)}* \frac{(x-3)(x+3)}{(y-4)(y+4)}= \frac{(y+4)(y+3)}{(x-3)(x+2)}* \frac{(x-3)(x+3)}{(y-4)(y+4)}= \\ \\ = \frac{(x+3)(y+3)}{(x+2)(y-4)}$

в)
$\frac{2ab^3}{5bc^2}: \frac{-8b^2}{10bc}= \frac{2ab^2}{5c^2}: \frac{-4b}{5c}= \\ \\ = \frac{2ab^2}{5c^2}* \frac{5c}{-4b}=- \frac{ab}{2c}$

г)
$\frac{x^2-9}{-6}: \frac{x-3}{18}= \frac{(x-3)(x+3)}{-6}* \frac{18}{x-3}=-3(x+3)=-3x-9$