Question

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

Answer 1

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)
$frac{(b-1)(b+1)}{(b-1)(3b-1)}* frac{3b-1}{b}+ frac{1}{b}= frac{b+1}{b}+ frac{1}{b} = frac{b+2}{b}$