Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1486

\[\boxed{\mathbf{1486}\mathbf{.}}\]

\[1)\ y = \sqrt{\frac{x^{2} - 6x - 16}{x^{2} - 12x + 11}}\]

\[x^{2} - 12x + 11 = 0\]

\[D = 144 - 44 = 100\]

\[x_{1} = \frac{12 - 10}{2} = 1;\ \]

\[x_{2} = \frac{12 + 10}{2} = 11;\]

\[(x - 1)(x - 11) = 0.\]

\[x^{2} - 6x - 16 = 0\]

\[D = 36 + 64 = 100\]

\[x_{1} = \frac{6 - 10}{2} = - 2;\]

\[x_{2} = \frac{6 + 10}{2} = 8;\]

\[(x + 2)(x - 8) = 0.\]

\[Получим:\]

\[y = \sqrt{\frac{(x + 2)(x - 8)}{(x - 1)(x - 11)}}.\]

\[Имеет\ смысл\ при:\]

\[(x + 2)(x - 1)(x - 8)(x - 11) \geq 0\]

\[x \leq - 2\ \ \ \]

\[1 < x \leq 8\]

\[x > 11.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 2\rbrack \cup (1;\ 8\rbrack \cup (11;\ + \infty).\]

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

\[x - 3 > 0\]

\[x > 3.\]

\[Имеет\ смысл\ при:\]

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

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

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

\[x - 3 \leq \frac{1}{2}\]

\[x \leq 3\frac{1}{2}.\]

\[Ответ:\ \ x \in \left( 3;\ 3\frac{1}{2} \right\rbrack.\]