11/14:22/7=11/14:44/14=11/14*14/44=1/4
ОДЗ
7-3x>0⇒3x<7⇒x<7/3
7-3x≠1⇒3x≠6⇒x≠2
9x²-6x+1>0⇒(3x-1)²>0⇒3x-1≠0⇒3x≠1⇒x≠1/3
x∈(-∞;1/3) U (1/3;2) U (2;2 1/3)
1)x∈(-∞;1/3) U (1/3;2)
9x²-6x+1≥(7-3x)²
(3x-1)²-(7-3x)²≥0
(3x-1-7+3x)(3x-1+7-3x)≥0
(6x-8)*6≥0
6x-8≥0
6x≥8
x≥1 1/3
x∈[1 1/3;2)
2)x∈(2;2 1/3)
x≤1 1/3 решения нет
Ответ x∈[1 1/3;2)
=90 там просто под коренное число в квадрате равно этому же числу без корня
![log_{ \frac{1}{3} }(4x-3) - log_{ \frac{1}{3} } (1-x) \geq -2](https://tex.z-dn.net/?f=+log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D%284x-3%29+-+log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%281-x%29+%5Cgeq+-2)
ОДЗ:
![\left \{ {{4x-3\ \textgreater \ 0} \atop {1-x\ \textgreater \ 0}} \right. , \left \{ {{x\ \textgreater \ 0,75} \atop {x\ \textless \ 1}} \right. , =\ \textgreater \ 0,75\ \textless \ x\ \textless \ 1](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B4x-3%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7B1-x%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%2C++++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+0%2C75%7D+%5Catop+%7Bx%5C+%5Ctextless+%5C+1%7D%7D+%5Cright.+%2C++++++%3D%5C+%5Ctextgreater+%5C+++++++++0%2C75%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+1)
![log_{ \frac{1}{3} } \frac{4x-3}{1-x} \geq -2](https://tex.z-dn.net/?f=+log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%5Cfrac%7B4x-3%7D%7B1-x%7D+%5Cgeq+-2+)
![-2= log_{ \frac{1}{3} } ( \frac{1}{3} ) ^{-2} = log_{ \frac{1}{3} } 9](https://tex.z-dn.net/?f=-2%3D+log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%28+%5Cfrac%7B1%7D%7B3%7D+%29+%5E%7B-2%7D+%3D++log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+9)
![log_{ \frac{1}{3} } \frac{4x-3}{1-x} \geq log_{ \frac{1}{3} } 9](https://tex.z-dn.net/?f=+log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%5Cfrac%7B4x-3%7D%7B1-x%7D++%5Cgeq++log_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+9)
основание логарифма а=1/3. 0<1/3<1
знак неравенства меняем
![\frac{4x-3}{1-x} \leq 9, \frac{4x-3-9*(1-x)}{1-x} \leq 0, \frac{13x-12}{1-x} \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B4x-3%7D%7B1-x%7D++%5Cleq+9%2C++++++%5Cfrac%7B4x-3-9%2A%281-x%29%7D%7B1-x%7D+%5Cleq+0%2C+++++%5Cfrac%7B13x-12%7D%7B1-x%7D++%5Cleq+0+)
метод интервалов:
13х-12=0, 1-x≠0
x=12/13, x≠1
- + -
-----------[12/13]--------------(1)--------------->x
x≤12/13. x>1
включая ОДЗ, получим:
x∈(0,75; 12/13]
Ответ:
Объяснение:
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