Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 25

\[\boxed{\mathbf{25.}}\]

\[\textbf{а)}\log_{\frac{\sqrt{10}}{3}}{(1 - 3x)} < 2\]

\[\log_{\frac{\sqrt{10}}{3}}{(1 - 3x)} < \log_{\frac{\sqrt{10}}{3}}\left( \frac{10}{9} \right)\]

\[1 - 3x > 0\]

\[3x < 1\]

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

\[M = \left( - \infty;\frac{1}{3} \right).\]

\[1 - 3x < \frac{10}{9}\]

\[- 3x < \frac{1}{9}\]

\[x > - \frac{1}{27}.\]

\[Решение\ неравенства:\]

\[x \in \left( - \frac{1}{27};\frac{1}{3} \right).\]

\[Ответ:\ x \in \left( - \frac{1}{27};\frac{1}{3} \right).\]

\[\textbf{б)}\log_{\frac{\sqrt{6}}{3}}{(2x - 1)} > 2\]

\[\log_{\frac{\sqrt{6}}{3}}{(2x - 1)} > \log_{\frac{\sqrt{6}}{3}}\left( \frac{2}{3} \right)\]

\[2x - 1 > 0\]

\[2x > 1\]

\[x > 0,5.\]

\[M = (0,5; + \infty).\]

\[2x - 1 < \frac{2}{3}\]

\[2x < \frac{5}{3}\]

\[x < \frac{5}{6}.\]

\[Решение\ неравенства:\]

\[x \in \left( \frac{1}{2};\frac{5}{6} \right).\]

\[Ответ:\ x \in \left( \frac{1}{2};\ \frac{5}{6} \right).\]

\[\textbf{в)}\log_{0,5}{(x^{2} - 1)} > - 2\]

\[\log_{0,5}{(x^{2} - 1)} > \log_{0,5}{(4)}\]

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

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

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

\[M = ( - \infty; - 1) \cup (1; + \infty).\]

\[x^{2} - 1 < 4\]

\[x^{2} < 5\]

\[- \sqrt{5} < x < \sqrt{5}.\]

\[Решение\ неравенства:\]

\[x \in \left( - \infty; - \sqrt{5} \right) \cup \left( 1;\sqrt{5} \right).\]

\[Ответ:\ x \in \left( - \infty; - \sqrt{5} \right) \cup \left( 1;\sqrt{5} \right).\]

\[\textbf{г)}\log_{0,5}\left( x^{2} + 1 \right) < - 1\]

\[\log_{0,5}\left( x^{2} + 1 \right) < \log_{0,5}(2)\]

\[x^{2} + 1 > 0\]

\[x = R.\]

\[M = R.\]

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

\[x^{2} > 1\]

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

\[Решение\ неравенства:\]

\[x \in ( - \infty; - 1) \cup (1; + \infty).\]

\[Ответ:\ x \in ( - \infty; - 1) \cup (1; + \infty).\]