Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 397

\[\boxed{\mathbf{397}\mathbf{.}}\]

\[1)\ 4\log_{4}x - 33\log_{x}4 \leq 1\]

\[4\log_{4}x - 33 \bullet \frac{\log_{4}4}{\log_{4}x} - 1 \leq 0\]

\[4\log_{4}x - \frac{33}{\log_{4}x} - 1 \leq 0\]

\[Пусть\ y = \log_{4}x:\]

\[4y - \frac{33}{y} - 1 \leq 0\]

\[\frac{4y^{2} - y - 33}{y} \leq 0\]

\[D = 1^{2} + 4 \bullet 4 \bullet 33 = 1 + 528 =\]

\[= 529\]

\[y_{1} = \frac{1 - 23}{2 \bullet 4} = - \frac{22}{8} = - \frac{11}{4};\]

\[y_{2} = \frac{1 + 23}{2 \bullet 4} = \frac{24}{8} = 3.\]

\[\frac{\left( y + \frac{11}{4} \right)(y - 3)}{y} \leq 0\]

\[\left( y + \frac{11}{4} \right) \bullet y \bullet (y - 3) \leq 0\]

\[y \leq - \frac{11}{4};\text{\ \ }0 < y \leq 3.\]

\[1)\ \log_{4}x \leq - \frac{11}{4}\]

\[\log_{4}x \leq \log_{4}4^{- \frac{11}{4}}\]

\[x \leq 4^{- \frac{11}{4}}\ \]

\[x \leq \sqrt[4]{4^{- 11}}.\]

\[2)\ \log_{4}x > 0\]

\[\log_{4}x > \log_{4}1\]

\[x > 1.\]

\[3)\ \log_{4}x \leq 3\]

\[\log_{4}x \leq \log_{4}4^{3}\]

\[x \leq 4^{3}\ \]

\[x \leq 64.\]

\[имеет\ смысл\ при:\]

\[x > 0;\text{\ \ }x \neq 1.\]

\[Ответ:\ \ 0 < x \leq \sqrt[4]{4^{- 11}};\ \ \]

\[1 < x \leq 64.\]

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

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

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

\[Пусть\ y = \log_{3}x:\]

\[\frac{1}{y} \leq 4(1 - y)\]

\[\frac{1 - 4y(1 - y)}{y} \leq 0\]

\[\frac{1 - 4y + 4y^{2}}{y} \leq 0\]

\[\frac{(1 - 2y)^{2}}{y} \leq 0\]

\[y < 0\ \ и\ \ y = \frac{1}{2}.\]

\[1)\ \log_{3}x < 0\]

\[\log_{3}x < \log_{3}1\ \]

\[x < 1.\]

\[2)\ \log_{3}x = \frac{1}{2}\]

\[\log_{3}x = \log_{3}3^{\frac{1}{2}}\]

\[x = 3^{\frac{1}{2}} = \sqrt{3}.\]

\[имеет\ смысл\ при:\]

\[x > 0;\ x \neq 1.\]

\[Ответ:\ \ 0 < x < 1;\ \ x = \sqrt{3}.\]