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МатематикаРешебник по алгебре и начала математического анализа 11 класс Колягин Задание 924
Задание 924
\[1)\log_{6x + 1}(25x) - 2\log_{25x}(6x + 1) > 1\]
\[\log_{6x + 1}(25x) - \frac{2}{\log_{6x + 1}(25x)} - 1 > 0\]
\[y = \log_{6x + 1}(25x):\]
\[y - \frac{2}{y} - 1 > 0\]
\[\frac{y^{2} - y - 2}{y} > 0\]
\[D = 1 + 8 = 9\]
\[y_{1} = \frac{1 - 3}{2} = - 1;\]
\[y_{2} = \frac{1 + 3}{2} = 2;\]
\[\frac{(y + 1)(y - 2)}{y} > 0\]
\[- 1 < y < 0;\ \ \ y > 2.\]
\[Область\ определения:\]
\[x > 0;\ \ x \neq \frac{1}{25};\]
\[6x + 1 > 1.\]
\[1)\ - 1 < \log_{6x + 1}{25x} < 0\]
\[(6x + 1)^{- 1} < 25x < (6x + 1)^{0}\]
\[\frac{1}{6x + 1} < 25x < 1\]
\[\frac{25x(6x + 1) - 1}{6x + 1} > 0;\ x < \frac{1}{25}\]
\[\frac{150x^{2} + 25x - 1}{6x + 1} > 0\]
\[D = 625 + 600 = 1225\]
\[x_{1} = \frac{- 25 - 35}{2 \bullet 150} = - \frac{1}{5};\]
\[x_{2} = \frac{- 25 + 35}{2 \bullet 150} = \frac{1}{30};\]
\[\frac{\left( x + \frac{1}{5} \right)\left( x - \frac{1}{30} \right)}{6x + 1} > 0\]
\[- \frac{1}{5} < x < - \frac{1}{6};\ \ \ \]
\[x > \frac{1}{30}.\]
\[2)\ \log_{6x + 1}(25x) > 2\]
\[25x > (6x + 1)^{2}\]
\[25x > 36x^{2} + 12x + 1\]
\[36x^{2} - 13x + 1 < 0\]
\[D = 169 - 144 = 25\]
\[x_{1} = \frac{13 - 5}{2 \bullet 36} = \frac{1}{9};\]
\[x_{2} = \frac{13 + 5}{2 \bullet 36} = \frac{1}{4};\]
\[\left( x - \frac{1}{9} \right)\left( x - \frac{1}{4} \right) < 0\]
\[\frac{1}{9} < x < \frac{1}{4}.\]
\[Ответ:\ \ x \in \left( \frac{1}{30};\ \frac{1}{25} \right) \cup \left( \frac{1}{9};\ \frac{1}{4} \right).\]
\[2)\log_{6x - 1}\frac{x}{6x - 1} > 2\log_{x}(6x - 1)\]
\[\log_{6x - 1}x - \log_{6x - 1}(6x - 1) > \frac{2}{\log_{6x - 1}x}\]
\[y = \log_{6x - 1}x:\]
\[y - 1 > \frac{2}{y}\]
\[\frac{y^{2} - y - 2}{y} > 0\]
\[D = 1 + 8 = 9\]
\[y_{1} = \frac{1 - 3}{2} = - 1;\]
\[y_{2} = \frac{1 + 3}{2} = 2;\]
\[\frac{(y + 1)(y - 2)}{y} > 0\]
\[- 1 < y < 0;\ \ \ y > 2.\]
\[Область\ определения:\]
\[x > 0;\ \ \ x \neq 1;\]
\[x > \frac{1}{6};\ \ \ x \neq \frac{1}{3}.\]
\[\frac{1}{6} < x < \frac{1}{3}:\]
\[- 1 < \log_{6x - 1}x < 0\]
\[(6x - 1)^{0} < x < (6x - 1)^{- 1}\]
\[1 < x < \frac{1}{6x - 1}.\]
\[\log_{6x - 1}x > 2\]
\[x < (6x - 1)^{2}\]
\[x < 36x^{2} - 12x + 1\]
\[36x^{2} - 13x + 1 > 0\]
\[D = 169 - 144 = 25\]
\[x_{1} = \frac{13 - 5}{2 \bullet 36} = \frac{1}{9};\]
\[x_{2} = \frac{13 + 5}{2 \bullet 36} = \frac{1}{4};\]
\[\left( x - \frac{1}{9} \right)\left( x - \frac{1}{4} \right) > 0\]
\[x < \frac{1}{9};\ \ \ \ x > \frac{1}{4}.\]
\[\frac{1}{3} < x < 1:\]
\[\log_{6x - 1}x > 2\]
\[x > 36x^{2} - 12x + 1\]
\[x < 1.\]
\[- 1 < \log_{6x - 1}x < 0\]
\[\frac{1}{6x - 1} < x < 1\]
\[\frac{x(6x - 1) - 1}{6x - 1} > 0\]
\[\frac{6x^{2} - x - 1}{6x - 1} > 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{1 - 5}{2 \bullet 6} = - \frac{1}{3};\]
\[x_{2} = \frac{1 + 5}{2 \bullet 6} = \frac{1}{2};\]
\[\frac{\left( x + \frac{1}{3} \right)\left( x - \frac{1}{2} \right)}{6x - 1} > 0\]
\[- \frac{1}{3} < x < \frac{1}{6};\ \ \ x > \frac{1}{2}.\]
\[Ответ:\ \ x \in \left( \frac{1}{4};\ \frac{1}{3} \right) \cup \left( \frac{1}{2};\ 1 \right).\]
\[3)\log_{x + 4}\left( \sqrt{x + 5} + 1 \right) \leq 1\]
\[Область\ определения:\]
\[x > - 4;\ x \neq - 3;\ x \geq - 5.\]
\[x > - 3:\]
\[\sqrt{x + 5} + 1 \leq x + 4\]
\[\sqrt{x + 5} \leq x + 3\]
\[x + 5 \leq x^{2} + 6x + 9\]
\[x^{2} + 5x + 4 \geq 0\]
\[D = 25 - 16 = 9\]
\[x_{1} = \frac{- 5 - 3}{2} = - 4;\]
\[x_{2} = \frac{- 5 + 3}{2} = - 1;\]
\[(x + 4)(x + 1) \geq 0\]
\[x \leq - 4;\ \ \ x \geq - 1.\]
\[x < - 3:\]
\[\sqrt{x + 5} + 1 \geq x + 4\]
\[- 4 \leq x \leq 1.\]
\[Ответ:\ \]
\[x \in ( - 4;\ - 3) \cup \lbrack - 1;\ + \infty).\]
