Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 917

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

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

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

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

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

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

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

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

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

\[= 0\]

\[1)\ \log_{2}x - 1 = 0\]

\[\log_{2}x = 1\]

\[\log_{2}x = \log_{2}2\]

\[x = 2.\]

\[2)\ \log_{2}(x - 3) - 1 = 0\]

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

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

\[x - 3 = 2\ \]

\[x = 5.\]

\[имеет\ смысл\ при:\]

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

\[x - 3 > 0 \Longrightarrow x > 3.\]

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

\[x(x - 3) > 0\]

\[x < 0\ \ и\ \ x > 3.\]

\[Ответ:\ \ x = 5.\]