Question

Решите номер 33.Есть вложение.

Answer 1

$frac{16a^2-8a+1}{1-4a+x-4ax} = frac{(4a-1)^2}{x(1-4a)+1-4a} = frac{(4a-1)^2}{(1-4a)(x+1)} =frac{1-4a}{x+1}$

$frac{1-6c+y-6cy}{1-12c+36c^2} = frac{(1-6c)+y(1-6c)}{(1-6c)^2} = frac{(1-6c)(1+y)}{(1-6c)^2} = frac{1+y}{1-6c}$