Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1598

\[\boxed{\mathbf{1598}\mathbf{.}}\]

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

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

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

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

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

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

\[D = 25 - 16 = 9\]

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

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

\[Имеет\ смысл\ при:\]

\[x > 0;\text{\ \ }x \neq 1;\]

\[5 - x > 0\]

\[x < 5.\]

\[Ответ:\ \ x = 4.\]

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

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

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

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

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

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

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

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

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

\[D = 9 + 216 = 225\]

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

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

\[Имеет\ смысл\ при:\]

\[3 - x > 0 \rightarrow x < 3;\]

\[3 - x \neq 1 \rightarrow x \neq 2;\]

\[7 - x > 0 \rightarrow x < 7.\]

\[Ответ:\ \ x = - 9.\]