Помогите пожалуйста решить 4,03 б , 4,05в , 4,04 б

403.
= $= \frac{(b+4)^{2} }{b}* \frac{b}{(b+4)}( \frac{1}{b+4)} + \frac{1}{4-b})+ \frac{b}{b-4}=$ $(b+4)( \frac{1}{b+4)} + \frac{1}{4-b})+ \frac{b}{b-4}= \frac{b+4}{b+4} + \frac{b+4}{4-b}+ \frac{b}{b-4}=1- \frac{b+4}{b-4}+ \frac{b}{b-4}=$ $1+ \frac{b-b-4}{b-4}=1- \frac{4}{b-4}= \frac{b-4-4}{b-4}= \frac{b-8}{b-4}$

404
$(\frac{c(5c-1)}{ (5c-1)^{2} }- \frac{4}{(5c-1)(5c+1)}):( \frac{5c-1-3}{5c-1})- \frac{c}{5c+1}=$ $(\frac{c(5c+1)-4}{(5c-1)(5c+1)}):( \frac{5c-4}{5c-1})- \frac{c}{5c+1}=(\frac{5c^{2} +c-4}{(5c-1)(5c+1)})*( \frac{5c-1}{5c-4})- \frac{c}{5c+1}=$ $(\frac{5c^{2} +c-4}{5c+1})*( \frac{1}{5c-4})- \frac{c}{5c+1}=$ $(\frac{(c+1)(5c-4)}{5c+1})*( \frac{1}{5c-4})- \frac{c}{5c+1}=(\frac{c+1}{5c+1})- \frac{c}{5c+1}=(\frac{c+1-c}{5c+1})=\frac{1}{5c+1}$

405
3b²-4b+1=3b²-3b-b+1=3b(b-1)-(b-1)=(3b-1)(b-1)
$\frac{(b-1)(b+1)}{(b-1)(3b-1)}* \frac{3b-1}{b}+ \frac{1}{b}= \frac{b+1}{b}+ \frac{1}{b} = \frac{b+2}{b}$