ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 919

\[\boxed{\mathbf{919}.}\]

\[\log_{a}^{2}x^{2} > 1;\ \ x \neq 0\]

\[t = \log_{a}x:\]

\[t^{2} - 1 = 0\]

\[(t - 1)(t + 1) = 0\]

\[t = - 1;\ \ t = 1.\]

\[t < - 1;\ \ t > 1.\]

\[\left\{ \begin{matrix} \log_{a}x^{2} < - 1 \\ \log_{a}x^{2} > 1\ \ \ \ \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \log_{a}x^{2} < \log_{a}\frac{1}{a} \\ \log_{a}x^{2} > \log_{a}a \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ a < 0;\ \ \ a = 1 - нет\ \]

\[решений.\]

\[2)\ 0 < a < 1:\]

\[\left\{ \begin{matrix} \log_{a}\ x^{2} < \log_{a}\frac{1}{a} \\ \log_{a}\ x^{2} > \log_{a}\text{\ a} \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} > \frac{1}{a} \\ x^{2} < a \\ x \neq 0\ \\ \end{matrix} \right.\ \]

\[x^{2} - \frac{1}{a} > 0\]

\[\left( x - \frac{1}{\sqrt{a}} \right)\left( x + \frac{1}{\sqrt{a}} \right) > 0\]

\[x < - \frac{1}{\sqrt{a}};\ \ \ x > \frac{1}{\sqrt{a}}.\]

\[x^{2} - a < 0\]

\[\left( x - \sqrt{a} \right)\left( x + \sqrt{a} \right) < 0\]

\[- \sqrt{a} < x < \sqrt{a}\]

\[x \in \left( - \infty - \frac{1}{\sqrt{a}} \right) \cup \left( - \sqrt{a};0 \right) \cup\]

\[\cup \left( 0;\sqrt{a} \right) \cup \left( \frac{1}{\sqrt{a}}; + \infty \right).\]

\[3)\ a > 1:\]

\[\left\{ \begin{matrix} \log_{a}\ x^{2} < \log_{a}\frac{1}{a} \\ \log_{a}\ x^{2} > \log_{a}\text{\ a} \\ x \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x^{2} < \frac{1}{a} \\ x^{2} > a \\ x \neq 0\ \ \\ \end{matrix} \right.\ \]

\[x^{2} - \frac{1}{a} < 0\]

\[\left( x - \frac{1}{\sqrt{a}} \right)\left( x + \frac{1}{\sqrt{a}} \right) < 0\]

\[- \frac{1}{\sqrt{a}} < x < \frac{1}{\sqrt{a}}.\]

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

\[\left( x - \sqrt{a} \right)\left( x + \sqrt{a} \right) > 0\]

\[x < - \sqrt{a};\ \ x > \sqrt{a}\]

\[x \in \left( - \infty; - \sqrt{a} \right) \cup \left( - \frac{1}{\sqrt{a}};0 \right) \cup\]

\[\cup \left( 0;\frac{1}{\sqrt{a}} \right) \cup \left( \sqrt{a} + \infty \right)\text{.\ }\]

\[Ответ:при\ a < 0;a = 1 - нет\]

\[\ решений;\]

\[при\ 0 < a < 1 - x \in\]

\[\in \left( - \infty - \frac{1}{\sqrt{a}} \right) \cup \left( - \sqrt{a};0 \right) \cup\]

\[\cup \left( 0;\sqrt{a} \right) \cup \left( \frac{1}{\sqrt{a}}; + \infty \right);\]

\[при\ a > 1 - x \in \left( - \infty; - \sqrt{a} \right) \cup\]

\[\cup \left( - \frac{1}{\sqrt{a}};0 \right) \cup \left( 0;\frac{1}{\sqrt{a}} \right) \cup\]

\[\cup \left( \sqrt{a} + \infty \right).\]