ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 362

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\[1)\log_{\frac{1}{3}}{\log_{2}x^{2}} > 0\]

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

\[\log_{2}x^{2} < 1\]

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

\[x^{2} < 2\]

\[- \sqrt{2} < x < \sqrt{2}\]

\[имеет\ смысл\ при:\]

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

\[\log_{2}x^{2} > \log_{2}1\]

\[x^{2} > 1\]

\[x < - 1\ \ и\ \ x > 1.\]

\[2)\ x^{2} > 0 \Longrightarrow x \neq 0.\]

\[Ответ:\ \ - \sqrt{2} < x < - 1;\ \ \]

\[1 < x < \sqrt{2}.\]

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

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

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

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

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

\[x^{2} > 1 + \frac{1}{8}\]

\[x^{2} > \frac{9}{8}\]

\[- \frac{3}{2\sqrt{2}} < x < \frac{3}{2\sqrt{2}}\]

\[имеет\ смысл\ при:\]

\[1)\ \log_{\frac{1}{2}}\left( x^{2} - 1 \right) > 0\]

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

\[x^{2} - 1 < 1\]

\[x^{2} < 2\]

\[- \sqrt{2} < x < \sqrt{2}.\]

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

\[x^{2} > 1\]

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

\[Ответ:\ \ - \sqrt{2} < x < - \frac{3}{2\sqrt{2}};\ \ \]

\[\frac{3}{2\sqrt{2}} < x < \sqrt{2}.\]