ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 908

\[\boxed{\mathbf{908}.}\]

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

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

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

\[Пусть\ y = \log_{4}x:\]

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

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

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

\[= 1 + 528 = 529\]

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

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

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

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

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

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

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

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

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

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

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

\[x > 1.\]

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

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

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

\[x \leq 64.\]

\[имеет\ смысл\ при:\]

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

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

\[\ \ 1 < x \leq 64.\]

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

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

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

\[Пусть\ y = \log_{3}x:\]

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

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

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

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

\[y < 0\ \ и\ \ y = \frac{1}{2}.\]

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

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

\[x < 1.\]

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

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

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

\[имеет\ смысл\ при:\]

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

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

\[3)\log_{x + 2}{x²} > 1\ \]

\[Область\ определения:\]

\[x \neq 0;\ \ x > - 2.\]

\[\log_{x + 2}\ x^{2} > \log_{x + 2}\ (x + 2)\]

\[\left\{ \begin{matrix} x + 2 > 1\ \ \ \\ x^{2} > x + 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x > - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 1)(x - 2) > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x > - 1 \\ x < - 1 \\ x > 2\ \ \ \\ \end{matrix} \right.\ \rightarrow x > 2.\]

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

\[(x + 1)(x - 2) > 0.\]

\[\left\{ \begin{matrix} 0 < x + 2 < 1 \\ x^{2} < x + 2\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} - 2 < x < - 1\ \ \ \ \ \ \ \ \ \ \\ (x + 1)(x - 2) > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 2 < x < - 1 \\ - 1 < x < 2\ \ \ \\ \end{matrix} \right.\ \rightarrow нет\ решений.\]

\[Ответ:x > 2.\]

\[4)\log_{x^{2} + 2}{(3x + 6)} \leq 1\]

\[Область\ определения:\]

\[3x + 6 > 0\]

\[3x > - 6\]

\[x > - 2.\]

\[\log_{x^{2} + 2}\ (3x + 6) \leq\]

\[\leq \log_{x^{2} + 2}\ \left( x^{2} + 2 \right)\]

\[\left\{ \begin{matrix} x > - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2 \geq 3x + 6 \\ \end{matrix} \right.\ \]

\[x^{2} + 2 - 3x - 6 \geq 0\]

\[x^{2} - 3x - 4 \geq 0\]

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\ \ x_{2} = - 1\]

\[(x + 1)(x - 4) \geq 0\]

\[\left\{ \begin{matrix} x > - 2 \\ x \leq - 1 \\ x \geq 4\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} - 2 < x \leq - 1 \\ x \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ - 2 < x \leq - 1;\ \ x \geq - 4.\]