ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 914

\[1)\log_{\frac{1}{2}}\left( \log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \right) \leq 0\]

\[\log_{\frac{1}{2}}\left( \log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \right) \leq \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{0}\]

\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} \geq 1\]

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

\[\frac{3x + 1}{x - 1} \leq \frac{1}{2}\]

\[\frac{2(3x + 1) - (x - 1)}{2(x - 1)} \leq 0\]

\[\frac{6x + 2 - x + 1}{x - 1} \leq 0\]

\[\frac{5x + 3}{x - 1} \leq 0\]

\[- \frac{3}{5} \leq x < 1.\]

\[Область\ определения:\]

\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} > 0\]

\[\log_{\frac{1}{2}}\frac{3x + 1}{x - 1} > \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{0}\]

\[\frac{3x + 1}{x - 1} < 1,\ \ \ \frac{3x + 1}{x - 1} > 0\]

\[\frac{(3x + 1) - (x - 1)}{x - 1} < 0\]

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

\[- 1 < x < 1.\]

\[\frac{3x + 1}{x - 1} > 0\]

\[x < - \frac{1}{3};\ \ x > 1.\]

\[- 1 < x < - \frac{1}{3}.\]

\[Ответ:\ \ x \in \left\lbrack - \frac{3}{5};\ - \frac{1}{3} \right).\]

\[2)\log_{\frac{1}{3}}\left( \log_{4}\left( x^{2} - 5 \right) \right) > 0\]

\[\log_{\frac{1}{3}}\left( \log_{4}\left( x^{2} - 5 \right) \right) > \log_{\frac{1}{3}}\left( \frac{1}{3} \right)^{0}\]

\[\log_{4}\left( x^{2} - 5 \right) < 1\]

\[\log_{4}\left( x^{2} - 5 \right) < \log_{4}4^{1}\]

\[x^{2} - 5 < 4\]

\[x^{2} - 9 < 0\]

\[(x + 3)(x - 3) < 0\]

\[- 3 < x < 3.\]

\[Область\ определения:\]

\[\log_{4}\left( x^{2} - 5 \right) > 0\]

\[\log_{4}\left( x^{2} - 5 \right) > \log_{4}4^{0}\]

\[x^{2} - 5 > 1\]

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

\[\left( x + \sqrt{6} \right)\left( x - \sqrt{6} \right) > 0\]

\[x < - \sqrt{6};\ \ \ x > \sqrt{6}.\]

\[Ответ:\ \ \]

\[x \in \left( - 3;\ - \sqrt{6} \right) \cup \left( \sqrt{6};\ 3 \right).\]

\[\ \]