ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 404

\[\boxed{\mathbf{404.}}\]

\[1)\log_{\frac{1}{3}}\left( 2^{x + 2} - 4^{x} \right) \geq - 2\]

\[\log_{\frac{1}{3}}\left( 2^{x + 2} - 4^{x} \right) \geq \log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{- 2}\]

\[2^{x + 2} - 4^{x} \leq \left( \frac{1}{3} \right)^{- 2}\]

\[2^{x} \bullet 2^{2} - 2^{2x} \leq 3^{2}\]

\[4 \bullet 2^{x} - 2^{2x} \leq 9\]

\[2^{2x} - 4 \bullet 2^{x} + 9 \geq 0\]

\[Пусть\ y = 2^{x}:\]

\[y^{2} - 4y + 9 \geq 0\]

\[D = 4^{2} - 4 \bullet 9 = 16 - 36 =\]

\[= - 20 < 0\]

\[a = 1 > 0 \Longrightarrow \ y - любое\ число.\]

\[имеет\ смысл\ при:\]

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

\[2^{x} \bullet 2^{2} - 2^{2x} > 0\]

\[2^{x} \bullet \left( 4 - 2^{x} \right) > 0\]

\[4 - 2^{x} > 0\]

\[2^{x} < 4\]

\[2^{x} < 2^{2}\]

\[x < 2\]

\[Ответ:\ \ x < 2.\]

\[2)\log_{\frac{1}{\sqrt{5}}}\left( 6^{x + 1} - 36^{x} \right) \geq - 2\]

\[6^{x + 1} - 36^{x} \leq \left( \frac{1}{\sqrt{5}} \right)^{- 2}\]

\[6^{x} \bullet 6 - 6^{2x} \leq \left( \sqrt{5} \right)^{2}\]

\[6 \bullet 6^{x} - 6^{2x} \leq 5\]

\[6^{2x} - 6 \bullet 6^{x} + 5 \geq 0\]

\[Пусть\ y = 6^{x}:\]

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

\[D = 6^{2} - 4 \bullet 5 = 36 - 20 = 16\]

\[y_{1} = \frac{6 - 4}{2} = 1;\text{\ \ }y_{2} = \frac{6 + 4}{2} = 5.\]

\[(y - 1)(y - 5) \geq 0\]

\[y \leq 1\ \ и\ \ y \geq 5.\]

\[1)\ 6^{x} \leq 1\]

\[6^{x} \leq 6^{0}\]

\[x \leq 0.\]

\[2)\ 6^{x} \geq 5\]

\[\log_{6}6^{x} \geq \log_{6}5\ \]

\[x \geq \log_{6}5.\]

\[имеет\ смысл\ при:\]

\[6^{x + 1} - 36^{x} > 0\]

\[6^{x} \bullet 6 - 6^{2x} > 0\]

\[6^{x} \bullet \left( 6 - 6^{x} \right) > 0\]

\[6 - 6^{x} > 0\]

\[6^{x} < 6\]

\[x < 1.\]

\[Ответ:\ \ x \leq 0;\ \ \log_{6}5 \leq x < 1.\]

\[\text{\ \ }\]