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МатематикаГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 858
Задание 858
\[1)\ \sqrt{7 - \log_{\sqrt{3}}\left( 3x^{2} - 24x \right)} =\]
\[= \log_{9}\left( x^{2} - 8x \right);\text{\ \ \ }\]
\[\sin x < tg\ 2x;\]
\[\sqrt{7 - 2\log_{3}{3\left( x^{2} - 8x \right)}} =\]
\[= \frac{1}{2}\log_{3}\left( x^{2} - 8x \right)\]
\[2\sqrt{7 - \left( 2 + 2\log_{3}\left( x^{2} - 8x \right) \right)} =\]
\[= \log_{3}\left( x^{2} - 8x \right)\]
\[y = \log_{3}\left( x^{2} - 8x \right):\]
\[2\sqrt{7 - (2 + 2y)} = y\]
\[4(5 - 2y) = y^{2}\]
\[y^{2} + 8y - 20 = 0\]
\[D = 64 + 80 = 144\]
\[y_{1} = \frac{- 8 - 12}{2} = - 10;\]
\[y_{2} = \frac{- 8 + 12}{2} = 2.\]
\[Область\ определения:\]
\[\log_{3}\left( x^{2} - 8x \right) \geq 0.\]
\[Подставим:\]
\[\log_{3}\left( x^{2} - 8x \right) = 2\]
\[x^{2} - 8x = 3^{2}\]
\[x^{2} - 8x - 9 = 0\]
\[D = 64 + 36 = 100\]
\[x_{1} = \frac{8 - 10}{2} = - 1;\]
\[x_{2} = \frac{8 + 10}{2} = 9.\]
\[Ответ:\ - 1.\]
\[2)\ \sqrt{\log_{\frac{1}{6}}\left( x^{2} + \frac{1}{6}x \right)} \bullet \log_{36x^{2} + 6x}6 = 1;\text{\ \ \ }\]
\[\sin x > tg\ 6x;\]
\[\sqrt{- \log_{6}\left( x^{2} + \frac{1}{6}x \right)} = \frac{1}{\log_{36x^{2} + 6x}6}\]
\[\sqrt{- \log_{6}\left( x^{2} + \frac{1}{6}x \right)} = \log_{6}\left( 36x^{2} + 6x \right)\]
\[\sqrt{- \log_{6}\left( x^{2} + \frac{1}{6}x \right)} =\]
\[= \log_{6}36 + \log_{6}\left( x^{2} + \frac{1}{6}x \right)\]
\[y = \log_{6}\left( x^{2} + \frac{1}{6}x \right):\]
\[\sqrt{- y} = 2 + y\]
\[- y = 4 + 4y + y^{2}\]
\[y^{2} + 5y + 4 = 0\]
\[D = 25 - 16 = 9\]
\[y_{1} = \frac{- 5 - 3}{2} = - 4;\]
\[y_{2} = \frac{- 5 + 3}{2} = - 1.\]
\[Область\ определения:\]
\[- 2 \leq \log_{6}\left( x^{2} + \frac{1}{6}x \right) \leq 0.\]
\[Подставим:\]
\[\log_{6}\left( x^{2} + \frac{1}{6}x \right) = - 1\]
\[x^{2} + \frac{1}{6}x = 6^{- 1}\]
\[x^{2} + \frac{1}{6}x - \frac{1}{6} = 0\]
\[6x^{2} + x - 1 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{- 1 - 5}{2 \bullet 6} = - \frac{1}{2};\]
\[x_{2} = \frac{- 1 + 5}{2 \bullet 6} = \frac{1}{3}.\]
\[Ответ:\ \ \frac{1}{3}.\]
