ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 856

\[\boxed{\mathbf{856}.}\]

\[1 + \log_{6}\frac{x + 3}{x + 7} = \frac{1}{4}\log_{\sqrt{6}}(x - 1)^{2}\]

\[Область\ определения:\]

\[1)\ \ \frac{x + 3}{x + 7} > 0\]

\[x < - 7;\ \ x > - 3.\]

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

\[x - любое\ число,\ кроме\ x = 1.\]

\[D(y) =\]

\[= ( - \infty; - 7) \cup ( - 3;1) \cup (1;\ + \infty).\]

\[1 + \log_{6}\frac{x + 3}{x + 7} =\]

\[= \frac{1}{4} \cdot 2\log_{6}\ (x - 1)^{2}\]

\[1 + \log_{6}\frac{x + 3}{x + 7} = \frac{1}{2}\log_{6}\ (x - 1)^{2}\]

\[1 + \log_{6}\frac{x + 3}{x + 7} = \log_{6}\ (x - 1)^{2 \cdot \frac{1}{2}}\]

\[\log_{6}\frac{x + 3}{x + 7} - \log_{6}\ |x - 1| = - 1\]

\[\log_{6}\ \left( \frac{x + 3}{(x + 7)|x - 1|} \right) = - 1\]

\[\frac{x + 3}{(x + 7)|x - 1|} = 6^{- 1}\]

\[\frac{x + 3}{(x + 7)|x - 1|} = \frac{1}{6}\]

\[6 \cdot (x + 3) = (x + 7)|x - 1|\]

\[x > - 1:\]

\[6(x + 3) = (x + 7)(x - 1)\]

\[6x + 18 = x^{2} - x + 7x - 7\]

\[x^{2} = 25\]

\[x = \pm 5.\]

\[x < - 1:\]

\[6 \cdot (x + 3) = (x + 7)(1 - x)\]

\[6x + 18 = x - x^{2} + 7 - 7x\]

\[x^{2} + 12x + 11 = 0\]

\[D_{1} = 36 - 11 = 25\]

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

\[\text{\ \ }x_{2} = - 6 - 5 = - 11.\]

\[Ответ:x = - 11;x = - 1;x = 5.\]