О.Д.З.:
![\begin{cases} 2x+1\ \textgreater \ 0,\ 2x+1 \neq 1; \\ 5-2x\ \textgreater \ 0,\ 5-2x \neq 1; \\ 5+8x-4x^2\ \textgreater \ 0;\\ 1+4x+4x^2\ \textgreater \ 0;\end{cases}\ \Leftrightarrow \begin{cases} x\ \textgreater \ -0,5,\ x \neq 0; \\ x\ \textless \ 2,5,\ x \neq 2; \\ (x+0,5)(x-2,5)\ \textless \ 0;\\ (1+2x)^2\ \textgreater \ 0;\end{cases} \\ \Rightarrow \begin{cases} x \in (-0,5;0) \cup (0;2) \cup (2;2,5); \\ (x+0,5)(x-2,5)\ \textless \ 0;\\ x \neq 0,5;\end{cases} \Rightarrow \\ \Rightarrow \boxed {x \in (-0,5;0) \cup (0;2) \cup (2;2,5)}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D+2x%2B1%5C+%5Ctextgreater+%5C+0%2C%5C+2x%2B1+%5Cneq+1%3B+%5C%5C+5-2x%5C+%5Ctextgreater+%5C+0%2C%5C+5-2x+%5Cneq+1%3B+%5C%5C+5%2B8x-4x%5E2%5C+%5Ctextgreater+%5C+0%3B%5C%5C+1%2B4x%2B4x%5E2%5C+%5Ctextgreater+%5C+0%3B%5Cend%7Bcases%7D%5C+%5CLeftrightarrow+%5Cbegin%7Bcases%7D+x%5C+%5Ctextgreater+%5C+-0%2C5%2C%5C+x+%5Cneq+0%3B+%5C%5C+x%5C+%5Ctextless+%5C+2%2C5%2C%5C+x+%5Cneq+2%3B+%5C%5C+%28x%2B0%2C5%29%28x-2%2C5%29%5C+%5Ctextless+%5C+0%3B%5C%5C+%281%2B2x%29%5E2%5C+%5Ctextgreater+%5C+0%3B%5Cend%7Bcases%7D+%5C%5C%0A%5CRightarrow+%5Cbegin%7Bcases%7D+x+%5Cin+%28-0%2C5%3B0%29+%5Ccup+%280%3B2%29+%5Ccup+%282%3B2%2C5%29%3B+%5C%5C+%28x%2B0%2C5%29%28x-2%2C5%29%5C+%5Ctextless+%5C+0%3B%5C%5C+x+%5Cneq+0%2C5%3B%5Cend%7Bcases%7D+%5CRightarrow++%5C%5C+%5CRightarrow++%5Cboxed+%7Bx+%5Cin+%28-0%2C5%3B0%29+%5Ccup+%280%3B2%29+%5Ccup+%282%3B2%2C5%29%7D)
Преобразуем (формула перехода к новому основанию, напр. 10):
![\frac{lg(5-2x)(1+2x)}{lg(1+2x)} + \frac{lg(1+2x)^2}{lg(5-2x)} =4](https://tex.z-dn.net/?f=+%5Cfrac%7Blg%285-2x%29%281%2B2x%29%7D%7Blg%281%2B2x%29%7D+%2B+%5Cfrac%7Blg%281%2B2x%29%5E2%7D%7Blg%285-2x%29%7D+%3D4)
![\frac{lg(5-2x)}{lg(1+2x)} + \frac{lg(1+2x)}{lg(1+2x)}+ \frac{2lg(1+2x)}{lg(5-2x)} =4](https://tex.z-dn.net/?f=+%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%2B+%5Cfrac%7Blg%281%2B2x%29%7D%7Blg%281%2B2x%29%7D%2B+%5Cfrac%7B2lg%281%2B2x%29%7D%7Blg%285-2x%29%7D+%3D4)
![\frac{lg(5-2x)}{lg(1+2x)} + 1+2* \frac{lg(1+2x)}{lg(5-2x)} =4](https://tex.z-dn.net/?f=+%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%2B+1%2B2%2A+%5Cfrac%7Blg%281%2B2x%29%7D%7Blg%285-2x%29%7D+%3D4)
![\frac{lg(5-2x)}{lg(1+2x)} +2* \frac{lg(1+2x)}{lg(5-2x)} -3=0](https://tex.z-dn.net/?f=+%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%2B2%2A+%5Cfrac%7Blg%281%2B2x%29%7D%7Blg%285-2x%29%7D+-3%3D0)
Замена
![\frac{lg(5-2x)}{lg(1+2x)} =t](https://tex.z-dn.net/?f=+%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%3Dt)
![t+ \frac{2}{t} -3=0](https://tex.z-dn.net/?f=t%2B+%5Cfrac%7B2%7D%7Bt%7D+-3%3D0)
t² - 3t + 2 = 0
t=1 или t=2
![1)\ \frac{lg(5-2x)}{lg(1+2x)} =1](https://tex.z-dn.net/?f=1%29%5C++%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%3D1)
5 - 2x = 1 + 2x
4x = 4
x = 1 ∈ ОДЗ
![2)\ \frac{lg(5-2x)}{lg(1+2x)} =2](https://tex.z-dn.net/?f=2%29%5C++%5Cfrac%7Blg%285-2x%29%7D%7Blg%281%2B2x%29%7D+%3D2)
5 - 2x = (1 + 2x)²
5 - 2x = 1 + 4x + 4x²
2x² + 3x - 2 = 0
x = 0,5 ∈ ОДЗ
x = -2 ∉ ОДЗ.
Ответ: 0,5; 1.
БЫЛО 40
ПОКУПАТЕЛЕЙ 7
ПОКУПАЛИ ПО 4(ПИР)
ОСТАЛОСЬ ?
40-(7*4)=12 (ПИР)
ОТВЕТ 12
Ответ:16
Пошаговое объяснение:
5-литровых канистр было 22, значит 5*22=110 (литров) доставили в 5-литровых канистрах, остальные литры доставили в 15- литровых канистрах, узнаем сколько литров доставили в 15-литровых 200-110=90 литров доставили в 15-литровых канистрах. Узнаем сколько было 15-литровых канистр 90:15=6(канистр)
Как это можно не знать инет для чего?