Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 921

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Год:2020-2021-2022-2023
Тип:учебник

Задание 921

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

\[x^{2} - 4 > 0\]

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

\[x < - 2;\ \ \ x > 2.\]

\[\log_{0,5}x > 0\]

\[x < 1.\]

\[Область\ определения:\]

\[x > 0.\]

\[Ответ:\ \ x \in (1;\ 2).\]

\[2)\ (3x - 1) \bullet \log_{2}x > 0\]

\[3x - 1 > 0\]

\[3x > 1\]

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

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

\[x > 1.\]

\[Область\ определения:\]

\[x > 0.\]

\[Ответ:\ \ x \in \left( 0;\ \frac{1}{3} \right) \cup (1;\ + \infty).\]

\[3)\log_{7}{\log_{\frac{1}{3}}\frac{x^{2} + |x| - 30}{x + 6}} < 0\]

\[\log_{\frac{1}{3}}\frac{x^{2} + |x| - 30}{x + 6} < 1\]

\[\frac{x^{2} + |x| - 30}{x + 6} > \frac{1}{3}\]

\[\frac{3\left( x^{2} + |x| - 30 \right) - (x + 6)}{3(x + 6)} > 0\]

\[\frac{3x^{2} + \left( 3|x| - x \right) - 96}{x + 6} > 0\]

\[x \geq 0:\]

\[\frac{3x^{2} + 2x - 96}{x + 6} > 0\]

\[D = 4 + 1152 = 1156\]

\[x_{1} = \frac{- 2 - 34}{2 \bullet 3} = - 6;\]

\[\ x_{2} = \frac{- 2 + 34}{2 \bullet 3} = \frac{16}{3};\]

\[(x + 6)\left( x - \frac{16}{3} \right) > 0\]

\[x < - 6;\ \ \ x > \frac{16}{3}.\]

\[x \leq 0:\]

\[\frac{3x^{2} - 4x - 96}{x + 6} > 0\]

\[D = 16 + 1152 = 1168\]

\[x = \frac{4 \pm \sqrt{1168}}{2 \bullet 3} = \frac{4 \pm 2\sqrt{292}}{2 \bullet 3} =\]

\[= \frac{2 \pm \sqrt{292}}{3};\]

\[- 6 < x < \frac{2 - \sqrt{292}}{3}\]

\[x > \frac{2 + \sqrt{292}}{3}.\]

\[Область\ определения:\]

\[\log_{\frac{1}{3}}\frac{x^{2} + |x| - 30}{x + 6} > 0\]

\[\frac{x^{2} + |x| - 30}{x + 6} < 1\]

\[\frac{x^{2} + |x| - 30 - x - 6}{x + 6} < 0\]

\[\frac{x^{2} + \left( |x| - x \right) - 36}{x + 6} < 0.\]

\[x \geq 0:\]

\[\frac{x^{2} - 36}{x + 6} < 0\]

\[\frac{(x + 6)(x - 6)}{x + 6} < 0\]

\[x < 6;\ \ \ x \neq - 6.\]

\[x \leq 0:\]

\[\frac{x^{2} - 2x - 36}{x + 6} < 0\]

\[D = 4 + 144 = 148\]

\[x = \frac{2 \pm \sqrt{148}}{2} = \frac{2 \pm 2\sqrt{37}}{2} =\]

\[= 1 \pm \sqrt{37};\]

\[x < - 6;\]

\[1 - \sqrt{37} < x < 1 + \sqrt{37}.\]

\[Ответ:\ \ \]

\[x \in \left( 1 - \sqrt{37};\ \frac{2 - \sqrt{292}}{3} \right) \cup \left( \frac{16}{3};\ 6 \right).\]

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