![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]
0,73^2+0,27*0,73+0,27=0,73*(0,27+0,73)+0,27=0,73*1+0,27=0.73+0.27=1
- 3х/7 + 3 ( х - 2/7 ) + 11 = - 7/2 * х + 3 ( х + 1 )
(-3/7)х+3х-6/7+11=(-7/2)х+3х+3
(-3/7)х+(7/2)х=-8+6/7
(43/14)х=-7 1/7
х=(-50/7)*14/43
х=-100/43
х=-2 14/43