ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 875

\[\boxed{\mathbf{875}.}\]

\[1)\ \frac{1}{5 - \lg x} + \frac{2}{1 + \lg x} < 1\]

\[Пусть\ y = \lg x:\]

\[\frac{1}{5 - y} + \frac{2}{1 + y} < 1\]

\[\frac{y^{2} - 5y + 6}{(5 - y)(1 + y)} < 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[y_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }y_{2} = \frac{5 + 1}{2} = 3.\]

\[\frac{(y - 2)(y - 3)}{(5 - y)(1 + y)} < 0\]

\[(y + 1)(y - 2)(y - 3)(5 - y) < 0\]

\[(y + 1)(y - 2)(y - 3)(y - 5) > 0\]

\[y < - 1;\ \ 2 < y < 3;\ \ y > 5.\]

\[1)\ \lg x < - 1\]

\[\lg x < \lg 10^{- 1}\]

\[x < 10^{- 1}\]

\[\ x < 0,1.\]

\[2)\ \lg x > 2\]

\[\lg x > \lg 10^{2}\]

\[x > 10^{2}\]

\[x > 100.\]

\[3)\ \lg x < 3\]

\[\lg x < \lg 10^{3}\]

\[x < 10^{3}\]

\[x < 1000.\]

\[4)\ \lg x > 5\]

\[\lg x > \lg 10^{5}\]

\[x > 10^{5}\]

\[x > 100\ 000.\]

\[\ имеет\ смысл\ при:\ \]

\[x > 0.\]

\[Ответ:\ \ 0 < x < 0,1;\ \ \]

\[100 < x < 1000;\ \ x > 100\ 000.\]

\[2)\log_{3}\left( 2 - 3^{- x} \right) < x +\]

\[+ 1 - \log_{3}4\]

\[\log_{3}\left( 2 - 3^{- x} \right) < \log_{3}3^{x} +\]

\[+ \log_{3}3 - \log_{3}4\]

\[\log_{3}\left( 2 - 3^{- x} \right) < \log_{3}\frac{3^{x} \bullet 3}{4}\]

\[2 - 3^{- x} < \frac{3 \bullet 3^{x}}{4}\]

\[4\left( 2 - 3^{- x} \right) < 3 \bullet 3^{x}\]

\[8 - 4 \bullet 3^{- x} < 3 \bullet 3^{x}\]

\[3 \bullet 3^{x} - 8 + 4 \bullet 3^{- x} > 0\ \ \ \ \ | \bullet 3^{x}\]

\[3 \bullet 3^{2x} - 8 \bullet 3^{x} + 4 > 0\]

\[Пусть\ y = 3^{x}:\]

\[3y^{2} - 8y + 4 > 0\]

\[D = 8^{2} - 4 \bullet 3 \bullet 4 =\]

\[= 64 - 48 = 16\]

\[y_{1} = \frac{8 - 4}{2 \bullet 3} = \frac{4}{6} = \frac{2}{3};\]

\[y_{2} = \frac{8 + 4}{2 \bullet 3} = \frac{12}{6} = 2.\]

\[\left( y - \frac{2}{3} \right)(y - 2) > 0\]

\[y < \frac{2}{3};\ \ y > 2.\]

\[1)\ 3^{x} < \frac{2}{3}\]

\[\log_{3}3^{x} < \log_{3}\frac{2}{3}\ \]

\[x < \log_{3}\frac{2}{3}.\]

\[2)\ 3^{x} > 2\]

\[\log_{3}3^{x} > \log_{3}2\]

\[x > \log_{3}2.\]

\[имеет\ смысл\ при:\]

\[2 - 3^{- x} > 0\]

\[3^{- x} < 2\]

\[\log_{3}3^{- x} < \log_{3}2\]

\[- x < \log_{3}2\]

\[x > - \log_{3}2\]

\[x > \log_{3}2^{- 1}\]

\[x > \log_{3}{\frac{1}{2}.}\]

\[Ответ:\ \ \log_{3}\frac{1}{2} < x < \log_{3}\frac{2}{3};\]

\[\ \ x > \log_{3}2\text{.\ }\]

\[3)\log_{\frac{x - 1}{5x - 6}}\left( \sqrt{6} - 2x \right) < 0\]

\[\log_{\frac{x - 1}{5x - 6}}\left( \sqrt{6} - 2x \right) < \log_{\frac{x - 1}{5x - 6}}1\]

\[Основание\ логарифма:\]

\[\frac{x - 1}{5x - 6} > 1\]

\[\frac{x - 1}{5x - 6} - 1 > 0\]

\[\frac{x - 1 - 5x + 6}{5x - 6} > 0\]

\[\frac{5 - 4x}{5x - 6} > 0\]

\[\frac{4x - 5}{5x - 6} < 0\]

\[(5x - 6)(4x - 5) < 0\]

\[1,2 < x < 1,25.\]

\[1,2 < x < 1,25:\]

\[\sqrt{6} - 2x < 1\]

\[- 2x < 1 - \sqrt{6}\]

\[x > \frac{\sqrt{6} - 1}{2}.\]

\[x < 1,2\ или\ x > 1,25:\]

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

\[x < \frac{\sqrt{6} - 1}{2}.\]

\[имеет\ смысл\ при:\]

\[1)\ \frac{x - 1}{5x - 6} > 0\]

\[(x - 1)(5x - 6) > 0\]

\[x < 1;\text{\ \ }x > 1,2.\]

\[2)\ \sqrt{6} - 2x > 0\]

\[2x < \sqrt{6}\]

\[x < \frac{\sqrt{6}}{2}\]

\[Ответ:\ \ x < \frac{\sqrt{6} - 1}{2};\ \ \]

\[1,2 < x < \frac{\sqrt{6}}{2}.\]