Question

Помогите пожалуста срочно!!!!!заранее спасибо!!!!!!

Answer 1

$=\frac{10}{(5-b^2)(5+b^2)}+\frac{5-b^2}{(5-b^2)(5+b^2)}-\frac{5+b^2}{(5-b^2)(5+b^2)}=\frac{10+5-b^2-5-b^2}{(5-b^2)(5+b^2)}=\frac{10-2b^2}{(5-b^2)(5+b^2)}=\frac{2(5-b^2)}{(5-b^2)(5+b^2)}=\frac{2}{5+b^2}\\ 2>0; 5+b^2>0$