1) Область определения
![-log_{3}x\ \textgreater \ 0, log_{3}x\ \textless \ 0, x\ \textless \ 1, x\ \textgreater \ 0, 0\ \textless \ x\ \textless \ 1 ](https://tex.z-dn.net/?f=+-log_%7B3%7Dx%5C+%5Ctextgreater+%5C+0%2C++log_%7B3%7Dx%5C+%5Ctextless+%5C+0%2C++x%5C+%5Ctextless+%5C+1%2C+x%5C+%5Ctextgreater+%5C+0%2C+0%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+1%0A)
Обозначим:
![- log_{3} x=q,](https://tex.z-dn.net/?f=-+log_%7B3%7D+x%3Dq%2C)
тогда
![Log_{0,5}^2(q) - Log_{0,5}( q^{2} ) \leq 3, Log_{2^{-1}}^2(q) - Log_{2^{-1}}( q^{2} ) \leq 3,](https://tex.z-dn.net/?f=+Log_%7B0%2C5%7D%5E2%28q%29+-+Log_%7B0%2C5%7D%28+q%5E%7B2%7D+%29++%5Cleq+3%2C+Log_%7B2%5E%7B-1%7D%7D%5E2%28q%29+-+Log_%7B2%5E%7B-1%7D%7D%28+q%5E%7B2%7D+%29++%5Cleq+3%2C+)
![(-Log_{2}(q))^{2} +Log_{2}( q^{2} ) \leq 3, (Log_{2}(q))^{2} +2Log_{2}( q) \leq 3,](https://tex.z-dn.net/?f=%28-Log_%7B2%7D%28q%29%29%5E%7B2%7D+%2BLog_%7B2%7D%28+q%5E%7B2%7D+%29++%5Cleq+3%2C+%28Log_%7B2%7D%28q%29%29%5E%7B2%7D+%2B2Log_%7B2%7D%28+q%29++%5Cleq+3%2C)
![(Log_{2}(q))^{2} +2Log_{2}( q) -3 \leq 0, (Log_2(q) +3)(Log_2(q)-1) \leq 0,](https://tex.z-dn.net/?f=%28Log_%7B2%7D%28q%29%29%5E%7B2%7D+%2B2Log_%7B2%7D%28+q%29+-3+%5Cleq+0%2C+%28Log_2%28q%29+%2B3%29%28Log_2%28q%29-1%29+%5Cleq+0%2C)
рисуем интервалы
-∞___+____-3___-___1___+___+∞
![-3 \leq Log_{2}(q) \leq 1,](https://tex.z-dn.net/?f=-3+%5Cleq+Log_%7B2%7D%28q%29+%5Cleq+1%2C+)
1.
![- log_{3} x \geq \frac{1}{8} ,log_{3} x \leq - \frac{1}{8} ,x \leq 3^{- \frac{1}{8} }](https://tex.z-dn.net/?f=-+log_%7B3%7D+x+%5Cgeq++%5Cfrac%7B1%7D%7B8%7D+%2Clog_%7B3%7D+x+%5Cleq+-+%5Cfrac%7B1%7D%7B8%7D+%2Cx+%5Cleq+3%5E%7B-+%5Cfrac%7B1%7D%7B8%7D+%7D)
2.
Ответ:
![\frac{1}{9} \leq x \leq 3^{- \frac{1}{8}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B9%7D++%5Cleq+x+%5Cleq+3%5E%7B-+%5Cfrac%7B1%7D%7B8%7D)
2)
![Log_{|x-1|}(x-2)^2 \leq 2,](https://tex.z-dn.net/?f=Log_%7B%7Cx-1%7C%7D%28x-2%29%5E2+%5Cleq+2%2C)
Область определения:
![|x-1| \neq 0, |x-1| \neq 1, (x-2) \neq 0, x \neq 0, x \neq 1, x \neq 2](https://tex.z-dn.net/?f=%7Cx-1%7C+%5Cneq+0%2C++%7Cx-1%7C+%5Cneq+1%2C+%28x-2%29+%5Cneq++0%2C+x+%5Cneq+0%2C+x+%5Cneq+1%2C+x+%5Cneq+2)
получаем область определения: x∈(-∞;0)∪(0;1)∪(1;2)∪(2;+∞)
1. 0<|x-1|<1, x∈(0;1)∪(1;2) основание логарифма меньше 1,
![Log_{|x-1|}(x-2)^{2}\leq 2, Log_{|x-1|}(x-2)^{2} \leq Log_{|x-1|}(x-1)^{2},](https://tex.z-dn.net/?f=Log_%7B%7Cx-1%7C%7D%28x-2%29%5E%7B2%7D%5Cleq+2%2C+Log_%7B%7Cx-1%7C%7D%28x-2%29%5E%7B2%7D+%5Cleq+Log_%7B%7Cx-1%7C%7D%28x-1%29%5E%7B2%7D%2C)
![(x-2)^{2} \geq (x-1)^{2}](https://tex.z-dn.net/?f=%28x-2%29%5E%7B2%7D+%5Cgeq+%28x-1%29%5E%7B2%7D)
![x^2-4x+4 \qeq x^2-2x+1, 2x-3 \leq 0, x \leq 3/2](https://tex.z-dn.net/?f=x%5E2-4x%2B4%C2%A0%5Cqeq+x%5E2-2x%2B1%2C+2x-3+%5Cleq+0%2C+x+%5Cleq+3%2F2)
,
Учитывая условие x∈(0;1)∪(1;2), получаем : x∈(0;1)∪(1;3/2].
2. 1<|x-1|, x∈(-∞;0)∪(2;+∞), основание логарифма больше 1,
![Log_{|x-1|}(x-2)^{2}\leq 2,log_{|x-1|}(x-2)^{2} \leq log_{|x-1|}(x-1)^{2},](https://tex.z-dn.net/?f=Log_%7B%7Cx-1%7C%7D%28x-2%29%5E%7B2%7D%5Cleq+2%2Clog_%7B%7Cx-1%7C%7D%28x-2%29%5E%7B2%7D+%5Cleq+log_%7B%7Cx-1%7C%7D%28x-1%29%5E%7B2%7D%2C)
![(x-2)^{2}\leq (x-1)^{2}](https://tex.z-dn.net/?f=%28x-2%29%5E%7B2%7D%5Cleq+%28x-1%29%5E%7B2%7D)
Учитывая условие <span>x∈(-∞;0)∪(2;+∞)</span> , получаем: x∈(2;+∞).
ответ: x∈(0;1)∪(1;3/2]∪(2;+∞)
Sin²x-11sinx+10=0
Sin²x=a
a²-11a+10=0
Δ=81
√Δ=9
a1=(11-9)/2=2/2=1
a2=(11+9)2=20/2=20 не подходит sinx ∈ <-1;1>
sinx=1
x=π/2+2kπ k e C
1 - 1/2
2 - 1
3 - 0.25
4 - 8/11
5 - 14/13
6 - 18/7
Х:у=4 ,значит, у*4=х , у*4+у
_____ = 5у= 5* 4=20
_
у
у
Решение смотри во вложении.