Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 854

\[1)\ 1 + \log_{x}(5 - x) = \log_{7}4 \bullet \log_{x}7\]

\[\log_{x}x + \log_{x}(5 - x) = \frac{\log_{x}4}{\log_{x}7} \bullet \log_{x}7\]

\[\log_{x}{x(5 - x)} = \log_{x}4\]

\[x(5 - x) = 4\]

\[5x - x^{2} = 4\]

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

\[D = 25 - 16 = 9\]

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

\[x_{2} = \frac{5 + 3}{2} = 4.\]

\[Область\ определения:\]

\[x > 0;\ \ \ x \neq 1;\ \ \ x < 5.\]

\[Ответ:\ \ 4.\]

\[2)\ \left( \log_{9}(7 - x) + 1 \right)\log_{3 - x}3 = 1\]

\[\frac{1}{2}\log_{3}(7 - x) + \log_{3}3 = \frac{1}{\log_{3 - x}3}\]

\[\log_{3}(7 - x) + 2\log_{3}3 = 2\log_{3}(3 - x)\]

\[\log_{3}(7 - x) + \log_{3}9 = \log_{3}(3 - x)^{2}\]

\[\log_{3}{9(7 - x)} = \log_{3}(3 - x)^{2}\]

\[63 - 9x = 9 - 6x + x^{2}\]

\[x^{2} + 3x - 54 = 0\]

\[D = 9 + 216 = 225\]

\[x_{1} = \frac{- 3 - 15}{2} = - 9;\]

\[x_{2} = \frac{- 3 + 15}{2} = 6.\]

\[Область\ определения:\]

\[x < 3;\ \ \ x \neq 2;\ \ \ x < 7.\]

\[Ответ:\ - 9.\]