Question

Помогите пожалуйста даю 35 баллов

Answer 1

$\dfrac{(u+1)^2}{16-16u^3}:\dfrac{1-u^2}{(4u-4)^2}=\dfrac{(u+1)^2}{16(1-u^3)}:\dfrac{(1-u)(1+u)}{4^2(u-1)^2}=\\\\\\=\dfrac{(u+1)^2}{16\, (1-u)(1+u+u^2)}\cdot \dfrac{16\, (1-u)^2}{(1-u)(1+u)}=\dfrac{u+1}{1+u+u^2}$

Ответ:  №4 .