Помогите пожалуйста решить 4,03 б , 4,05в , 4,04 б. Created by Dmitry2314. algebra-ru

403.=404 4053b²-4b+1=3b²-3b-b+1=3b(b-1)-(b-1)=(3b-1)(b-1)

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

Dmitry2314 @Dmitry2314

October 2021 1 6 Report

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

Answers & Comments

DariosI

Verified answer

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

2 votes Thanks 1

Answers & Comments

Verified answer

vse sposoby polucheniya pentena nuzhny formuly

pervyj chlen vozrastayushej arifmeticheskoj progressii raven 02 najdite raznost p485f9e45c3e2bbb66ddc277fac4c228a 63288

najdite trehznachnoe polozhitelnoe chislo esli ego cifry obrazuyut geometricheskuyu701701566547c0667dfc596f1e263477 3512

рекомендуемые вопросы

Helpful Links

Helpful Social