ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 769

\[\boxed{\mathbf{769}\mathbf{.}}\]

\[1)\log_{6}\left( 49 - x^{2} \right)\]

\[49 - x^{2} > 0\]

\[x^{2} - 49 < 0\]

\[(x + 7)(x - 7) < 0\]

\[- 7 < x < 7\]

\[Ответ:\ - 7 < x < 7.\]

\[2)\log_{7}\left( x^{2} + x - 6 \right)\]

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

\[D = 1^{2} + 4 \bullet 6 = 1 + 24 = 25\]

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

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

\[(x + 3)(x - 2) > 0\]

\[x < - 3\ \ и\ \ x > 2\]

\[Ответ:\ \ x < - 3;\ \ x > 2.\]

\[3)\log_{\frac{1}{5}}\left( x^{2} + 2x + 7 \right)\]

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

\[D = 2^{2} - 4 \bullet 7 =\]

\[= 4 - 28 = - 24 < 0\]

\[a = 1 > 0\]

\[x - любое\ число\]

\[Ответ:\ \ при\ любом\ \text{x.}\]

\[4)\ y = \log_{0,3}\left( 7x - 2x^{2} - 3 \right)\]

\[7x - 2x^{2} - 3 > 0\]

\[2x^{2} - 7x + 3 = 0\]

\[D = 49 - 24 = 25\]

\[x_{1} = \frac{7 + 5}{4} = 3;\ \ \ \]

\[x_{2} = \frac{7 - 5}{4} = 0,5.\]

\[- 2 \cdot (x - 0,5)(x - 3) > 0\]

\[0,5 < x < 3.\]

\[Ответ:при\ 0,5 < x < 3.\]

\[5)\ y = \log_{2}\frac{2x - 1}{x - 5}\]

\[\frac{2x - 1}{x - 5} > 0\]

\[\frac{2 \cdot (x - 0,5)}{x - 5} > 0\]

\[x < 0,5;\ \ x > 5.\]

\[Ответ:при\ x < 0,5;\ \ x > 5.\]

\[6)\ y = \log_{\frac{1}{7}}\frac{5x + 1}{4 - x}\]

\[\frac{5x + 1}{4 - x} > 0\]

\[\frac{5 \cdot (x + 0,2)}{- (x - 4)} > 0\]

\[- 0,2 < x < 4.\]

\[Ответ:\ - 0,2 < x < 4.\]

\[7)\ y = \log_{10}\sqrt{x^{2} - 1}\]

\[\sqrt{x^{2} - 1} > 0\]

\[x^{2} - 1 > 0\]

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

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

\[Ответ:при\ x < - 1;x > 1.\]

\[8)\ y = \log_{0,1}\sqrt{9 - x^{2}}\]

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

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

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

\[- 3 < x < 3.\]

\[Ответ:при - 3 < x < 3.\]