ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 834

\(\boxed{\mathbf{834}.}\)

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

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

\[x_{1} + x_{2} = 3;\ \ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\ \ \ \ x_{2} = - 1;\]

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

\[x < - 1;\ \ x > 4\]

\[D(y) = ( - \infty; - 1) \cup (4; + \infty).\]

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

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

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

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = - 6\]

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

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

\[- 1 < x < 6\]

\[D(y) = ( - 1;6).\]

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

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

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

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

\[D(y) = ( - 5;\ - 3) \cup (3; + \infty).\]

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

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

\[x - 4 > 0\]

\[x > 4\]

\[D(y) = (4; + \infty).\]

\[5)\ y = \log_{\pi}{(2^{x} - 2)}\]

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

\[2^{x} > 2\]

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

\[x > 1\]

\[D(y) = (1; + \infty).\]

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

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

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

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

\[x - 1 > 2\]

\[x > 3\]

\[D(y) = (3; + \infty).\]