Помогите решить пожалуйста:)))

1)
$\frac{b+2}{4b^3-4b^2+b}- \frac{2-b}{1-8b^3}* \frac{4b^2+2b+1}{2b^2+b}= \\ \\ = \frac{b+2}{b(4b^2-4b+1)}- \frac{2-b}{(1-2b)(1+2b+4b^2)}* \frac{4b^2+2b+1}{b(2b+1)}= \\ \\ = \frac{b+2}{b(2b-1)^2}- \frac{2-b}{b(1-2b)(2b+1)}= \frac{b+2}{b(2b-1)^2}+ \frac{2-b}{b(2b-1)(2b+1)}= \\ \\ = \frac{(b+2)(2b+1)+(2-b)(2b-1)}{b(2b-1)^2(2b+1)}= \frac{2b^2+4b+b+2+4b-2b^2-2+b}{b(2b-1)^2(2b+1)}= \\ \\ = \frac{10b}{b(2b-1)^2(2b+1)}= \frac{10}{(2b-1)^2(2b+1)}$

2)
$(\frac{1}{1-2b} )^2 : \frac{10}{(2b-1)^2(1+2b)}=(- \frac{1}{2b-1} )^2* \frac{(2b-1)^2(1+2b)}{10} = \\ \\ = \frac{1}{(2b-1)^2}* \frac{(2b-1)^2(1+2b)}{10}= \frac{1+2b}{10}$

3)
$\frac{2-b}{5}+ \frac{1+2b}{10}= \frac{2(2-b)+1+2b}{10}= \frac{4-2b+1+2b}{10}= \frac{5}{10}= \frac{1}{2}=0.5$