Ответы
Ответ дал:
3
1) Разложим выражения под логарифмами на множители:
1.1)![5+9x-2x^{2}=0 5+9x-2x^{2}=0](https://tex.z-dn.net/?f=5%2B9x-2x%5E%7B2%7D%3D0)
![-2x^{2}+9x+5=0, D=81+4*5*2=121=11^{2} -2x^{2}+9x+5=0, D=81+4*5*2=121=11^{2}](https://tex.z-dn.net/?f=-2x%5E%7B2%7D%2B9x%2B5%3D0%2C+D%3D81%2B4%2A5%2A2%3D121%3D11%5E%7B2%7D)
![x_{1}= \frac{-9-11}{-4}=5 x_{1}= \frac{-9-11}{-4}=5](https://tex.z-dn.net/?f=x_%7B1%7D%3D+%5Cfrac%7B-9-11%7D%7B-4%7D%3D5)
![x_{2}= \frac{-9+11}{-4}=-0.5 x_{2}= \frac{-9+11}{-4}=-0.5](https://tex.z-dn.net/?f=x_%7B2%7D%3D+%5Cfrac%7B-9%2B11%7D%7B-4%7D%3D-0.5)
![-2x^{2}+9x+5=-2*(x+0.5)(x-5) -2x^{2}+9x+5=-2*(x+0.5)(x-5)](https://tex.z-dn.net/?f=-2x%5E%7B2%7D%2B9x%2B5%3D-2%2A%28x%2B0.5%29%28x-5%29)
1.2)![x^{2}-10x+25=(x-5)^{2} x^{2}-10x+25=(x-5)^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D-10x%2B25%3D%28x-5%29%5E%7B2%7D)
2) Найдем область допустимых значений (ОДЗ):
![5-x\ \textgreater \ 0 5-x\ \textgreater \ 0](https://tex.z-dn.net/?f=5-x%5C+%5Ctextgreater+%5C+0)
![5-x \neq 1 5-x \neq 1](https://tex.z-dn.net/?f=5-x+%5Cneq+1)
![-2*(x+0.5)(x-5)\ \textgreater \ 0 -2*(x+0.5)(x-5)\ \textgreater \ 0](https://tex.z-dn.net/?f=-2%2A%28x%2B0.5%29%28x-5%29%5C+%5Ctextgreater+%5C+0)
![1+2x\ \textgreater \ 0 1+2x\ \textgreater \ 0](https://tex.z-dn.net/?f=1%2B2x%5C+%5Ctextgreater+%5C+0)
![1+2x \neq 1 1+2x \neq 1](https://tex.z-dn.net/?f=1%2B2x+%5Cneq+1)
![(x-5)^{4}\ \textgreater \ 0 (x-5)^{4}\ \textgreater \ 0](https://tex.z-dn.net/?f=%28x-5%29%5E%7B4%7D%5C+%5Ctextgreater+%5C+0)
![x\ \textless \ 5 x\ \textless \ 5](https://tex.z-dn.net/?f=x%5C+%5Ctextless+%5C+5)
![x \neq 4 x \neq 4](https://tex.z-dn.net/?f=x+%5Cneq+4)
![-0.5\ \textless \ x\ \textless \ 5 -0.5\ \textless \ x\ \textless \ 5](https://tex.z-dn.net/?f=-0.5%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+5)
![x\ \textgreater \ -0.5 x\ \textgreater \ -0.5](https://tex.z-dn.net/?f=x%5C+%5Ctextgreater+%5C+-0.5)
![x \neq 0 x \neq 0](https://tex.z-dn.net/?f=x+%5Cneq+0)
Общее решение ОДЗ: x∈(-0.5;0)U(0;4)U(4;5)
3)![log_{5-x}(2*(x+0.5))+log_{5-x}(5-x)+log_{1+2x}(x-5)^{4} \leq 5 log_{5-x}(2*(x+0.5))+log_{5-x}(5-x)+log_{1+2x}(x-5)^{4} \leq 5](https://tex.z-dn.net/?f=log_%7B5-x%7D%282%2A%28x%2B0.5%29%29%2Blog_%7B5-x%7D%285-x%29%2Blog_%7B1%2B2x%7D%28x-5%29%5E%7B4%7D+%5Cleq+5)
![log_{5-x}(1+2x)+1+4log_{1+2x}(5-x) \leq 5 log_{5-x}(1+2x)+1+4log_{1+2x}(5-x) \leq 5](https://tex.z-dn.net/?f=log_%7B5-x%7D%281%2B2x%29%2B1%2B4log_%7B1%2B2x%7D%285-x%29+%5Cleq+5)
![log_{5-x}(1+2x)+4log_{1+2x}(5-x) \leq 4 log_{5-x}(1+2x)+4log_{1+2x}(5-x) \leq 4](https://tex.z-dn.net/?f=log_%7B5-x%7D%281%2B2x%29%2B4log_%7B1%2B2x%7D%285-x%29+%5Cleq+4)
Замена:![log_{1+2x}(5-x)=t log_{1+2x}(5-x)=t](https://tex.z-dn.net/?f=log_%7B1%2B2x%7D%285-x%29%3Dt)
![log_{5-x}(1+2x)= \frac{1}{log_{1+2x}(5-x)}=\frac{1}{t} log_{5-x}(1+2x)= \frac{1}{log_{1+2x}(5-x)}=\frac{1}{t}](https://tex.z-dn.net/?f=log_%7B5-x%7D%281%2B2x%29%3D+%5Cfrac%7B1%7D%7Blog_%7B1%2B2x%7D%285-x%29%7D%3D%5Cfrac%7B1%7D%7Bt%7D)
![\frac{1}{t}+4t \leq 4 \frac{1}{t}+4t \leq 4](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bt%7D%2B4t+%5Cleq+4)
![\frac{1+4t^{2}-4t}{t} \leq 0 \frac{1+4t^{2}-4t}{t} \leq 0](https://tex.z-dn.net/?f=%5Cfrac%7B1%2B4t%5E%7B2%7D-4t%7D%7Bt%7D+%5Cleq+0)
![4t^{2}-4t+1=(2t-1)^{2} 4t^{2}-4t+1=(2t-1)^{2}](https://tex.z-dn.net/?f=4t%5E%7B2%7D-4t%2B1%3D%282t-1%29%5E%7B2%7D)
t∈(-бесконечность; 0) - решение неравенства
4) Вернемся к замене:
![log_{1+2x}(5-x)\ \textless \ 0 log_{1+2x}(5-x)\ \textless \ 0](https://tex.z-dn.net/?f=log_%7B1%2B2x%7D%285-x%29%5C+%5Ctextless+%5C+0)
4.1)![\left \{ {{1+2x\ \textgreater \ 1} \atop {5-x\ \textless \ 1}} \right. \left \{ {{1+2x\ \textgreater \ 1} \atop {5-x\ \textless \ 1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B1%2B2x%5C+%5Ctextgreater+%5C+1%7D+%5Catop+%7B5-x%5C+%5Ctextless+%5C+1%7D%7D+%5Cright.)
![\left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ 4}} \right. \left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ 4}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7Bx%5C+%5Ctextgreater+%5C+4%7D%7D+%5Cright.)
- решение
4.2)![\left \{ {{0\ \textless \ 1+2x\ \textless \ 1} \atop {5-x\ \textgreater \ 1}} \right. \left \{ {{0\ \textless \ 1+2x\ \textless \ 1} \atop {5-x\ \textgreater \ 1}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7B0%5C+%5Ctextless+%5C+1%2B2x%5C+%5Ctextless+%5C+1%7D+%5Catop+%7B5-x%5C+%5Ctextgreater+%5C+1%7D%7D+%5Cright.)
![\left \{ {{-0.5\ \textless \ x\ \textless \ 0} \atop {x\ \textless \ 4}} \right. \left \{ {{-0.5\ \textless \ x\ \textless \ 0} \atop {x\ \textless \ 4}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7B-0.5%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+0%7D+%5Catop+%7Bx%5C+%5Ctextless+%5C+4%7D%7D+%5Cright.)
- решение
5) Сравним с ОДЗ, получим окончательное решение неравенства:
x∈(-0.5;0)U(4;5) - ответ
1.1)
1.2)
2) Найдем область допустимых значений (ОДЗ):
Общее решение ОДЗ: x∈(-0.5;0)U(0;4)U(4;5)
3)
Замена:
t∈(-бесконечность; 0) - решение неравенства
4) Вернемся к замене:
4.1)
4.2)
5) Сравним с ОДЗ, получим окончательное решение неравенства:
x∈(-0.5;0)U(4;5) - ответ
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