Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 406

\[\boxed{\mathbf{406.}}\]

\[\frac{1}{\log_{a}x - 1} + \frac{1}{\log_{a}x^{2} + 1} < - \frac{3}{2}\]

\[\frac{1}{\log_{a}x - 1} + \frac{1}{2\log_{a}x + 1} < - \frac{3}{2}\]

\[Пусть\ y = \log_{a}x:\]

\[\frac{1}{y - 1} + \frac{1}{2y + 1} < - \frac{3}{2}\]

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

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

\[\frac{6y + 6y^{2} + 3y - 6y - 3}{2(y - 1)(2y + 1)} < 0\]

\[\frac{6y^{2} + 3y - 3}{4(y - 1)(y + 0,5)} < 0\]

\[\frac{3 \bullet \left( 2y^{2} + y - 1 \right)}{4 \bullet (y - 1)(y + 0,5)} < 0\]

\[Разложим\ многочлен\ \]

\[на\ множители:\]

\[2y^{2} + y - 1 = 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

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

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{2}{4} = 0,5;\]

\[(y + 1)(y - 0,5) = 0.\]

\[Получим:\]

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

\[- 1 < y < - 0,5;\ \ \ 0,5 < y < 1.\]

\[1)\ - 1 < \log_{a}x < - \frac{1}{2}\]

\[\log_{a}a^{- 1} < \log_{a}x < \log_{a}a^{- \frac{1}{2}}\]

\[\ 0 < a < 1:\]

\[a^{- \frac{1}{2}} < x < a^{- 1}\]

\[\frac{1}{\sqrt{a}} < x < \frac{1}{a}.\]

\[\ a > 1:\]

\[a^{- 1} < x < a^{- \frac{1}{2}}\]

\[\frac{1}{a} < x < \frac{1}{\sqrt{a}}\text{..}\]

\[2)\ \frac{1}{2} < \log_{a}x < 1\]

\[\log_{a}a^{\frac{1}{2}} < \log_{a}x < \log_{a}a^{1}\]

\[0 < a < 1:\]

\[a^{1} < x < a^{\frac{1}{2}}\]

\[a < x < \sqrt{a}.\]

\[a > 1:\]

\[a^{\frac{1}{2}} < x < a^{1}\]

\[\sqrt{a} < x < a.\]

\[Ответ:\ \ если\ 0 < a < 1,\ \]

\[то\ \frac{1}{\sqrt{a}} < x < \frac{1}{a}\text{\ \ }и\ \ a < x < \sqrt{a};\]

\[если\ a > 1,\ то\ \frac{1}{a} < x < \frac{1}{\sqrt{a}}\text{\ \ }и\ \ \]

\[\sqrt{a} < x < a.\ \]