Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 331

\[\boxed{\mathbf{331}\mathbf{.}}\]

\[1)\ y = \log_{8}\left( x^{2} - 3x - 4 \right)\]

\[x^{2} - 3x - 4 > 0\]

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

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

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

\[(x + 1)(x - 4) > 0\]

\[x < - 1\ \ и\ \ x > 4\]

\[Ответ:\ \ x < - 1;\ \ x > 4.\]

\[2)\ y = \log_{\sqrt{3}}\left( - x^{2} + 5x + 6 \right)\]

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

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

\[D = 5^{2} + 4 \bullet 6 = 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{5 + 7}{2} = 6.\]

\[(x + 1)(x - 6) < 0\]

\[- 1 < x < 6\]

\[Ответ:\ \ - 1 < x < 6.\]

\[3)\ y = \log_{0,7}\frac{x^{2} - 9}{x + 5}\]

\[\frac{x^{2} - 9}{x + 5} > 0\]

\[\left( x^{2} - 9 \right)(x + 5) > 0\]

\[(x + 5)(x + 3)(x - 3) > 0\]

\[- 5 < x < - 3\ \ и\ \ x > 3\]

\[Ответ:\ \ - 5 < x < - 3;\ \ x > 3.\]

\[4)\ y = \log_{\frac{1}{3}}\frac{x - 4}{x^{2} + 4}\]

\[\frac{x - 4}{x^{2} + 4} > 0\]

\[x - 4 > 0\ \]

\[x > 4.\]

\[Ответ:\ \ x > 4.\]

\(5)\ y = \log_{\pi}\left( 2^{x} - 2 \right)\)

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

\[2^{x} > 2\]

\[2^{x} > 2^{1}\ \]

\[x > 1\]

\[Ответ:\ \ x > 1.\]

\[6)\ y = \log_{3}\left( 3^{x - 1} - 9 \right)\]

\[3^{x - 1} - 9 > 0\]

\[3^{x - 1} > 9\]

\[3^{x - 1} > 3^{2}\]

\[x - 1 > 2\ \]

\[x > 3\]

\[Ответ:\ \ x > 3.\]