ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 364

\[\boxed{\mathbf{364}\mathbf{.}}\]

\[1)\log_{0,2}^{2}x - 5\log_{0,2}x < - 6\]

\[\log_{0,2}^{2}x - 5\log_{0,2}x + 6 < 0\]

\[Пусть\ y = \log_{0,2}x:\]

\[y^{2} - 5y + 6 < 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }y_{2} = \frac{5 + 1}{2} = 3.\]

\[(y - 2)(y - 3) < 0\]

\[2 < y < 3.\]

\[1)\ \log_{0,2}x > 2\]

\[\log_{\frac{1}{5}}x > \log_{\frac{1}{5}}\left( \frac{1}{5} \right)^{2}\]

\[x < \left( \frac{1}{5} \right)^{2}\]

\[x < \frac{1}{25}.\]

\[2)\ \log_{0,2}x < 3\]

\[\log_{\frac{1}{5}}x < \log_{\frac{1}{5}}\left( \frac{1}{5} \right)^{3}\]

\[x > \left( \frac{1}{5} \right)^{3}\]

\[x > \frac{1}{125}.\]

\[имеет\ смысл\ при:\]

\[x > 0.\]

\[Ответ:\ \ \frac{1}{125} < x < \frac{1}{25}.\]

\[2)\log_{0,1}^{2}x + 3\log_{0,1}x > 4\]

\[\log_{0,1}^{2}x + 3\log_{0,1}x - 4 > 0\]

\[Пусть\ y = \log_{0,1}x:\]

\[y^{2} + 3y - 4 > 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 - 5}{2} = - 4;\ \]

\[y_{2} = \frac{- 3 + 5}{2} = 1.\]

\[(y + 4)(y - 1) > 0\]

\[y < - 4;\text{\ \ }y > 1.\]

\[1)\ \log_{0,1}x < - 4\]

\[\log_{0,1}x < \log_{0,1}(0,1)^{- 4}\]

\[x > (0,1)^{- 4}\ \]

\[x > 10\ 000.\]

\[2)\ \log_{0,1}x > 1\]

\[\log_{0,1}x > \log_{0,1}{0,1}\ \]

\[x < 0,1.\]

\[имеет\ смысл\ при:\]

\[x > 0.\]

\[Ответ:\ \ 0 < x < 0,1;\ \]

\[\ x > 10\ 000.\]