ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 357

\[\boxed{\mathbf{357}\mathbf{.}}\]

\[\log_{15}\left( (x - 3)(x - 5) \right) < \log_{15}15\]

\[(x - 3)(x - 5) < 15\]

\[x^{2} - 5x - 3x + 15 < 15\]

\[x^{2} - 8x < 0\]

\[x(x - 8) < 0\]

\[0 < x < 8.\]

\[имеет\ смысл\ при:\]

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

\[x - 5 > 0 \Longrightarrow x > 5.\]

\[Ответ:\ \ 5 < x < 8.\]

\[(x - 2)(12 - x) \leq \left( \frac{1}{3} \right)^{- 2}\]

\[12x - x^{2} - 24 + 2x \leq 3^{2}\]

\[- x^{2} + 14x - 24 \leq 9\]

\[x^{2} - 14x + 33 \geq 0\]

\[D = 14^{2} - 4 \bullet 33 =\]

\[= 196 - 132 = 64\]

\[x_{1} = \frac{14 - 8}{2} = 3;\text{\ \ }\]

\[x_{2} = \frac{14 + 8}{2} = 11.\]

\[(x - 3)(x - 11) \geq 0\]

\[x \leq 3;\text{\ \ }x \geq 11.\]

\[имеет\ смысл\ при:\]

\[x - 2 > 0 \Longrightarrow x > 2;\]

\[12 - x > 0 \Longrightarrow x < 12.\]

\[Ответ:\ \ 2 < x \leq 3;\ \ 11 \leq x < 12.\]