![3\cdot 9^x-10\cdot 3^x+3 \geq 0](https://tex.z-dn.net/?f=3%5Ccdot+9%5Ex-10%5Ccdot+3%5Ex%2B3+%5Cgeq+0)
Рассмотрим функцию
![f(x)=3\cdot 9^x-10\cdot 3^x+3](https://tex.z-dn.net/?f=f%28x%29%3D3%5Ccdot+9%5Ex-10%5Ccdot+3%5Ex%2B3+)
Область определения:
![D(f)=\mathbb{R}](https://tex.z-dn.net/?f=D%28f%29%3D%5Cmathbb%7BR%7D)
Приравниваем функцию к нулю:
![f(x)=0;\,\,\,\, 3\cdot 9^x-10\cdot 3^x+3 =0](https://tex.z-dn.net/?f=f%28x%29%3D0%3B%5C%2C%5C%2C%5C%2C%5C%2C+3%5Ccdot+9%5Ex-10%5Ccdot+3%5Ex%2B3+%3D0)
Сделаем замену. Пусть
![3^x=t](https://tex.z-dn.net/?f=3%5Ex%3Dt)
, <u><em>причем</em></u><em> </em>
![t\ \textgreater \ 0](https://tex.z-dn.net/?f=t%5C+%5Ctextgreater+%5C+0)
, тогда получаем:
![3\cdot t^2-10t+3=0](https://tex.z-dn.net/?f=3%5Ccdot+t%5E2-10t%2B3%3D0)
<em>Решая квадратное уравнение, имеем такие корни: </em>
![t_1= \frac{1}{3} ;\,\,\, t_2=3](https://tex.z-dn.net/?f=t_1%3D+%5Cfrac%7B1%7D%7B3%7D+%3B%5C%2C%5C%2C%5C%2C+t_2%3D3)
<em> Обратная замена:
</em>
![\left[\begin{array}{ccc}3^x=\frac{1}{3}\\ 3^x=3\end{array}\right\Rightarrow \left[\begin{array}{ccc}x_1=-1\\ x_2=1\end{array}\right](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5Ex%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C+3%5Ex%3D3%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D-1%5C%5C+x_2%3D1%5Cend%7Barray%7D%5Cright)
<em>Определим решение данного неравенства:
</em>___+___[-1]____-___[1]__+____
<u><em>Окончательный ответ:</em></u><em> </em>![x \in (-\infty;-1]\cup[1;+\infty)](https://tex.z-dn.net/?f=x+%5Cin+%28-%5Cinfty%3B-1%5D%5Ccup%5B1%3B%2B%5Cinfty%29)
![2.\,\, \log_{0.5}^2x+2\log_{0.5}x-3\ \textgreater \ 0](https://tex.z-dn.net/?f=2.%5C%2C%5C%2C+%5Clog_%7B0.5%7D%5E2x%2B2%5Clog_%7B0.5%7Dx-3%5C+%5Ctextgreater+%5C+0)
<em>ОДЗ: </em>
![x\ \textgreater \ 0](https://tex.z-dn.net/?f=x%5C+%5Ctextgreater+%5C+0)
<em>Представим левую часть в виде:
</em>
![(\log_{0.5}x+1)^2-4\ \textgreater \ 0\\ (\log_{0.5}x+1)^2\ \textgreater \ 4\\\\ \left[\begin{array}{ccc}\log_{0.5}x+1\ \textgreater \ 2\\ \log_{0.5}x+1\ \textless \ -2\end{array}\right\Rightarrow \left[\begin{array}{ccc}\log_{0.5}x\ \textgreater \ 1\\ \log_{0.5}x\ \textless \ -3\end{array}\right\Rightarrow \left[\begin{array}{ccc}\log_{0.5}x\ \textgreater \ \log_{0.5}0.5\\ \log_{0.5}x\ \textless \ \log_{0.5}0.5^{-3}\end{array}\right](https://tex.z-dn.net/?f=%28%5Clog_%7B0.5%7Dx%2B1%29%5E2-4%5C+%5Ctextgreater+%5C+0%5C%5C+%28%5Clog_%7B0.5%7Dx%2B1%29%5E2%5C+%5Ctextgreater+%5C+4%5C%5C%5C%5C++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_%7B0.5%7Dx%2B1%5C+%5Ctextgreater+%5C+2%5C%5C+%5Clog_%7B0.5%7Dx%2B1%5C+%5Ctextless+%5C+-2%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_%7B0.5%7Dx%5C+%5Ctextgreater+%5C+1%5C%5C+%5Clog_%7B0.5%7Dx%5C+%5Ctextless+%5C+-3%5Cend%7Barray%7D%5Cright%5CRightarrow++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_%7B0.5%7Dx%5C+%5Ctextgreater+%5C+%5Clog_%7B0.5%7D0.5%5C%5C+%5Clog_%7B0.5%7Dx%5C+%5Ctextless+%5C+%5Clog_%7B0.5%7D0.5%5E%7B-3%7D%5Cend%7Barray%7D%5Cright)
<u><em>Поскольку основание </em></u>
![0\ \textless \ 0.5\ \textless \ 1](https://tex.z-dn.net/?f=0%5C+%5Ctextless+%5C+0.5%5C+%5Ctextless+%5C+1)
<u><em>, функция убывающая, то знак неравенства меняется на противоположный.</em></u><em />
<em>То есть
</em>
![\left[\begin{array}{ccc}x\ \textless \ 0.5\\ x\ \textgreater \ 0.5^{-3}\end{array}\right](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C+%5Ctextless+%5C+0.5%5C%5C+x%5C+%5Ctextgreater+%5C+0.5%5E%7B-3%7D%5Cend%7Barray%7D%5Cright)
<em>С учетом ОДЗ: </em>
<u><em>Окончательный ответ:</em></u><em> </em>![x \in(0;0.5)\cup(8;+\infty)](https://tex.z-dn.net/?f=x+%5Cin%280%3B0.5%29%5Ccup%288%3B%2B%5Cinfty%29)
![3.\,\, 64^x+7\cdot 8^x-8 \leq 0\\ 8^{2x}+7\cdot 8^x-8 \leq 0](https://tex.z-dn.net/?f=3.%5C%2C%5C%2C+64%5Ex%2B7%5Ccdot+8%5Ex-8+%5Cleq+0%5C%5C+8%5E%7B2x%7D%2B7%5Ccdot+8%5Ex-8+%5Cleq+0)
<em>Выделим полный квадрат в левой части неравенства:
</em>
<u><em>Окончательный ответ</em></u><em> </em>![x \in (-\infty;0]](https://tex.z-dn.net/?f=x+%5Cin+%28-%5Cinfty%3B0%5D)