Question

номер

2 (б,г,ж)
.
Срочно пожалуйста:)

Answer 1

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