15/7x-10/7=x+2
15/7-x=2+10/7
8/7x=24/7
x=3
![\left\{\begin{matrix}x^{2}-x>0\\ x^{2}-x<2\end{matrix}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%5E%7B2%7D-x%3E0%5C%5C%20x%5E%7B2%7D-x%3C2%5Cend%7Bmatrix%7D%5Cright.)
Решим каждое неравенство по отдельности.
Первое:
![x^{2}-x>0\\x(x-1)>0\\\begin{bmatrix}\left\{\begin{matrix}x>0\\x-1>0\end{matrix}\right.\\\left\{\begin{matrix}x<0\\x-1<0\end{matrix}\right.\end{matrix}\\\\\begin{bmatrix}\left\{\begin{matrix}x>0\\x>1\end{matrix}\right.\\\left\{\begin{matrix}x<0\\x<1\end{matrix}\right.\end{matrix}\\\\\begin{bmatrix}x\in(1;+\infty)\\ x\in(-\infty;0)\end{matrix}\\x\in(-\infty;0)\cup(1;+\infty)](https://tex.z-dn.net/?f=x%5E%7B2%7D-x%3E0%5C%5Cx%28x-1%29%3E0%5C%5C%5Cbegin%7Bbmatrix%7D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3E0%5C%5Cx-1%3E0%5Cend%7Bmatrix%7D%5Cright.%5C%5C%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3C0%5C%5Cx-1%3C0%5Cend%7Bmatrix%7D%5Cright.%5Cend%7Bmatrix%7D%5C%5C%5C%5C%5Cbegin%7Bbmatrix%7D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3E0%5C%5Cx%3E1%5Cend%7Bmatrix%7D%5Cright.%5C%5C%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3C0%5C%5Cx%3C1%5Cend%7Bmatrix%7D%5Cright.%5Cend%7Bmatrix%7D%5C%5C%5C%5C%5Cbegin%7Bbmatrix%7Dx%5Cin%281%3B%2B%5Cinfty%29%5C%5C%20x%5Cin%28-%5Cinfty%3B0%29%5Cend%7Bmatrix%7D%5C%5Cx%5Cin%28-%5Cinfty%3B0%29%5Ccup%281%3B%2B%5Cinfty%29)
Второе:
![x^{2}-x<2\\x^{2}-x-2<0\\x^{2}+x-2x-2<0\\x(x+1)-2(x+1)<0\\(x+1)(x-2)<0\\\begin{bmatrix}\left\{\begin{matrix}x+1<0\\x-2>0\end{matrix}\right.\\ \left\{\begin{matrix}x+1>0\\x-2<0\end{matrix}\right.\end{matrix}\\\\\begin{bmatrix}\left\{\begin{matrix}x<-1\\x>2\end{matrix}\right.\\ \left\{\begin{matrix}x>-1\\x<2\end{matrix}\right.\end{matrix}\\\\\begin{bmatrix}x\in\O\\x\in(-1;2)\end{matrix}\\x\in(-1;2)](https://tex.z-dn.net/?f=x%5E%7B2%7D-x%3C2%5C%5Cx%5E%7B2%7D-x-2%3C0%5C%5Cx%5E%7B2%7D%2Bx-2x-2%3C0%5C%5Cx%28x%2B1%29-2%28x%2B1%29%3C0%5C%5C%28x%2B1%29%28x-2%29%3C0%5C%5C%5Cbegin%7Bbmatrix%7D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%2B1%3C0%5C%5Cx-2%3E0%5Cend%7Bmatrix%7D%5Cright.%5C%5C%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%2B1%3E0%5C%5Cx-2%3C0%5Cend%7Bmatrix%7D%5Cright.%5Cend%7Bmatrix%7D%5C%5C%5C%5C%5Cbegin%7Bbmatrix%7D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3C-1%5C%5Cx%3E2%5Cend%7Bmatrix%7D%5Cright.%5C%5C%20%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%3E-1%5C%5Cx%3C2%5Cend%7Bmatrix%7D%5Cright.%5Cend%7Bmatrix%7D%5C%5C%5C%5C%5Cbegin%7Bbmatrix%7Dx%5Cin%5CO%5C%5Cx%5Cin%28-1%3B2%29%5Cend%7Bmatrix%7D%5C%5Cx%5Cin%28-1%3B2%29)
Находим общее решение для системы неравенств:
![\left\{\begin{matrix}x^{2}-x>0\\ x^{2}-x<2\end{matrix}\right.\\\\\left\{\begin{matrix}x\in(-\infty;0)\cup(1;+\infty)\\x\in(-1;2)\end{matrix}\right.\\x\in(-1;0)\cup(1;2)](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%5E%7B2%7D-x%3E0%5C%5C%20x%5E%7B2%7D-x%3C2%5Cend%7Bmatrix%7D%5Cright.%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Bmatrix%7Dx%5Cin%28-%5Cinfty%3B0%29%5Ccup%281%3B%2B%5Cinfty%29%5C%5Cx%5Cin%28-1%3B2%29%5Cend%7Bmatrix%7D%5Cright.%5C%5Cx%5Cin%28-1%3B0%29%5Ccup%281%3B2%29)
2x+y=12 |×2 4x+2y=24
7x-2y=31 7x-2y=31
Суммируем эти уравнения:
11x=55 |÷11
x=5
2*5+y=12
10+y=12
y=2 ⇒ (5;2)
Ответ: в первой четверти.
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;+∞)
0.7=
![\frac{7}{10}](https://tex.z-dn.net/?f=+%5Cfrac%7B7%7D%7B10%7D+)
0.(15)=0+15*
![\frac{1}{99}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B99%7D+)
=
![\frac{5}{33}](https://tex.z-dn.net/?f=+%5Cfrac%7B5%7D%7B33%7D+)
1.2(5)=1.2(5)*10=12+0.(5)=12+5*
![\frac{1}{9}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B9%7D+)
=12+
![\frac{5}{9}](https://tex.z-dn.net/?f=+%5Cfrac%7B5%7D%7B9%7D+)
=
![\frac{113}{90}](https://tex.z-dn.net/?f=+%5Cfrac%7B113%7D%7B90%7D+)