Question

Помогите!!!!!!!!!!!!!!!​

Answer 1

Ответ:

$\frac{b^2+30b-169d^2+225}{3b-39d+45} = \frac{b^2+30b+225-169d^2}{3(b+15-13d)} = \frac{(b+15)^2-169d^2}{3(b+15-13d)} = \frac{(b+15-13d)(b+15+13d)}{3(b+15-13d)} = \frac{b+15+13d}{3} = \frac{b+13d+15}{3}$

A = b +13 d +15

B = 3