ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 924

\[1)\log_{6x + 1}(25x) - 2\log_{25x}(6x + 1) > 1\]

\[\log_{6x + 1}(25x) - \frac{2}{\log_{6x + 1}(25x)} - 1 > 0\]

\[y = \log_{6x + 1}(25x):\]

\[y - \frac{2}{y} - 1 > 0\]

\[\frac{y^{2} - y - 2}{y} > 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2} = - 1;\]

\[y_{2} = \frac{1 + 3}{2} = 2;\]

\[\frac{(y + 1)(y - 2)}{y} > 0\]

\[- 1 < y < 0;\ \ \ y > 2.\]

\[Область\ определения:\]

\[x > 0;\ \ x \neq \frac{1}{25};\]

\[6x + 1 > 1.\]

\[1)\ - 1 < \log_{6x + 1}{25x} < 0\]

\[(6x + 1)^{- 1} < 25x < (6x + 1)^{0}\]

\[\frac{1}{6x + 1} < 25x < 1\]

\[\frac{25x(6x + 1) - 1}{6x + 1} > 0;\ x < \frac{1}{25}\]

\[\frac{150x^{2} + 25x - 1}{6x + 1} > 0\]

\[D = 625 + 600 = 1225\]

\[x_{1} = \frac{- 25 - 35}{2 \bullet 150} = - \frac{1}{5};\]

\[x_{2} = \frac{- 25 + 35}{2 \bullet 150} = \frac{1}{30};\]

\[\frac{\left( x + \frac{1}{5} \right)\left( x - \frac{1}{30} \right)}{6x + 1} > 0\]

\[- \frac{1}{5} < x < - \frac{1}{6};\ \ \ \]

\[x > \frac{1}{30}.\]

\[2)\ \log_{6x + 1}(25x) > 2\]

\[25x > (6x + 1)^{2}\]

\[25x > 36x^{2} + 12x + 1\]

\[36x^{2} - 13x + 1 < 0\]

\[D = 169 - 144 = 25\]

\[x_{1} = \frac{13 - 5}{2 \bullet 36} = \frac{1}{9};\]

\[x_{2} = \frac{13 + 5}{2 \bullet 36} = \frac{1}{4};\]

\[\left( x - \frac{1}{9} \right)\left( x - \frac{1}{4} \right) < 0\]

\[\frac{1}{9} < x < \frac{1}{4}.\]

\[Ответ:\ \ x \in \left( \frac{1}{30};\ \frac{1}{25} \right) \cup \left( \frac{1}{9};\ \frac{1}{4} \right).\]

\[2)\log_{6x - 1}\frac{x}{6x - 1} > 2\log_{x}(6x - 1)\]

\[\log_{6x - 1}x - \log_{6x - 1}(6x - 1) > \frac{2}{\log_{6x - 1}x}\]

\[y = \log_{6x - 1}x:\]

\[y - 1 > \frac{2}{y}\]

\[\frac{y^{2} - y - 2}{y} > 0\]

\[D = 1 + 8 = 9\]

\[y_{1} = \frac{1 - 3}{2} = - 1;\]

\[y_{2} = \frac{1 + 3}{2} = 2;\]

\[\frac{(y + 1)(y - 2)}{y} > 0\]

\[- 1 < y < 0;\ \ \ y > 2.\]

\[Область\ определения:\]

\[x > 0;\ \ \ x \neq 1;\]

\[x > \frac{1}{6};\ \ \ x \neq \frac{1}{3}.\]

\[\frac{1}{6} < x < \frac{1}{3}:\]

\[- 1 < \log_{6x - 1}x < 0\]

\[(6x - 1)^{0} < x < (6x - 1)^{- 1}\]

\[1 < x < \frac{1}{6x - 1}.\]

\[\log_{6x - 1}x > 2\]

\[x < (6x - 1)^{2}\]

\[x < 36x^{2} - 12x + 1\]

\[36x^{2} - 13x + 1 > 0\]

\[D = 169 - 144 = 25\]

\[x_{1} = \frac{13 - 5}{2 \bullet 36} = \frac{1}{9};\]

\[x_{2} = \frac{13 + 5}{2 \bullet 36} = \frac{1}{4};\]

\[\left( x - \frac{1}{9} \right)\left( x - \frac{1}{4} \right) > 0\]

\[x < \frac{1}{9};\ \ \ \ x > \frac{1}{4}.\]

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

\[\log_{6x - 1}x > 2\]

\[x > 36x^{2} - 12x + 1\]

\[x < 1.\]

\[- 1 < \log_{6x - 1}x < 0\]

\[\frac{1}{6x - 1} < x < 1\]

\[\frac{x(6x - 1) - 1}{6x - 1} > 0\]

\[\frac{6x^{2} - x - 1}{6x - 1} > 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 - 5}{2 \bullet 6} = - \frac{1}{3};\]

\[x_{2} = \frac{1 + 5}{2 \bullet 6} = \frac{1}{2};\]

\[\frac{\left( x + \frac{1}{3} \right)\left( x - \frac{1}{2} \right)}{6x - 1} > 0\]

\[- \frac{1}{3} < x < \frac{1}{6};\ \ \ x > \frac{1}{2}.\]

\[Ответ:\ \ x \in \left( \frac{1}{4};\ \frac{1}{3} \right) \cup \left( \frac{1}{2};\ 1 \right).\]

\[3)\log_{x + 4}\left( \sqrt{x + 5} + 1 \right) \leq 1\]

\[Область\ определения:\]

\[x > - 4;\ x \neq - 3;\ x \geq - 5.\]

\[x > - 3:\]

\[\sqrt{x + 5} + 1 \leq x + 4\]

\[\sqrt{x + 5} \leq x + 3\]

\[x + 5 \leq x^{2} + 6x + 9\]

\[x^{2} + 5x + 4 \geq 0\]

\[D = 25 - 16 = 9\]

\[x_{1} = \frac{- 5 - 3}{2} = - 4;\]

\[x_{2} = \frac{- 5 + 3}{2} = - 1;\]

\[(x + 4)(x + 1) \geq 0\]

\[x \leq - 4;\ \ \ x \geq - 1.\]

\[x < - 3:\]

\[\sqrt{x + 5} + 1 \geq x + 4\]

\[- 4 \leq x \leq 1.\]

\[Ответ:\ \]

\[x \in ( - 4;\ - 3) \cup \lbrack - 1;\ + \infty).\]