Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1487

\[\boxed{\mathbf{1487}\mathbf{.}}\]

\[1)\ y = \sqrt{\log_{0,8}\left( x^{2} - 5x + 7 \right)}\]

\[x^{2} - 5x + 7 > 0\]

\[D = 25 - 49 = - 24 < 0\]

\[a > 0;\]

\[x - любое\ число.\]

\[Имеет\ смысл\ при:\]

\[\log_{0,8}\left( x^{2} - 5x + 7 \right) \geq 0\]

\[\log_{0,8}\left( x^{2} - 5x + 7 \right) \geq \log_{0,8}{0,8}^{0}\]

\[x^{2} - 5x + 7 \leq 1\]

\[x^{2} - 5x + 6 \leq 0\]

\[D = 25 - 24 = 1\]

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

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

\[(x - 2)(x - 3) \leq 0\]

\[2 \leq x \leq 3.\]

\[Ответ:\ \ x \in \lbrack 2;\ 3\rbrack.\]

\[2)\ y = \sqrt{\log_{0,5}\left( x^{2} - 9 \right)}\]

\[Имеет\ смысл\ при:\]

\[x^{2} - 9 > 0\]

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

\[x < - 3\ и\ x > 3.\]

\[\log_{0,5}\left( x^{2} - 9 \right) \geq 0\]

\[\log_{0,5}\left( x^{2} - 9 \right) \geq \log_{0,5}(0,5)^{0}\]

\[x^{2} - 9 \leq 1\]

\[x^{2} \leq 10\]

\[- \sqrt{10} \leq x \leq \sqrt{10}.\]

\[Ответ:\ \ \]

\[x \in \left\lbrack - \sqrt{10};\ - 3 \right) \cup \left( 3;\ \sqrt{10} \right\rbrack.\]