Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 857

\[\boxed{\mathbf{857}.}\]

\[1)\log_{2}x + \log_{x}4 = 3;\ \ \]

\[\ \ x > 0;\ \ \ x \neq 1\]

\[\log_{2}x + \log_{x}2^{2} = 3\]

\[\log_{2}x + 2\log_{x}2 = 3\]

\[\log_{2}x + 2 \cdot \frac{1}{\log_{2}x} = 3\]

\[t = \log_{2}x:\]

\[t + \frac{2}{t} - 3 = 0\]

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

\[t_{1} + t_{2} = 3;\ \ \ \ t_{1} \cdot t_{2} = 2\]

\[t_{1} = 2;\ \ t_{2} = 1.\]

\[\textbf{а)}\ \log_{2}x = 2\]

\[x = 2^{2} = 4.\]

\[\textbf{б)}\ \log_{2}x = 1\]

\[x = 2^{1} = 2.\]

\[Ответ:x = 2;x = 4.\]

\[2)\log_{2x - 1}(2x - 3) =\]

\[= \log_{2x - 3}{(2x - 1)}\ \]

\[Область\ определения:\]

\[2x - 1 > 0;\ \ \ \ 2x - 3 > 0\]

\[2x > 1;\ \ \ \ \ \ \ \ \ \ \ \ 2x > 3\]

\[x > 0,5;\ \ \ \ \ \ \ \ \ \ x > 1,5\]

\[x > 1,5;\ \ \ x \neq 2.\]

\[\log_{2x - 1}\ (2x - 3) =\]

\[= \frac{1}{\log_{2x - 1}(2x - 3)}\]

\[\log_{2x - 1}\ (2x - 3) = t:\]

\[t = \frac{1}{t}\]

\[t^{2} = 1\]

\[t = \pm 1.\]

\[\textbf{а)}\ \log_{2x - 1}\ (2x - 3) = 1\]

\[2x - 3 = (2x - 1)^{1}\]

\[2x - 3 = 2x - 1\]

\[- 3 = - 1\]

\[нет\ корней.\]

\[\textbf{б)}\ \log_{2x - 1}\ (2x - 3) = - 1\]

\[2x - 3 = (2x - 1)^{- 1}\]

\[2x - 3 = \frac{1}{2x - 1}\]

\[(2x - 3)(2x - 1) = 1\]

\[4x^{2} - 6x - 2x + 3 = 1\]

\[4x^{2} - 8x + 2 = 0\ \ \ \ \ \ \ \ |\ :2\]

\[2x^{2} - 4x + 1 = 0\]

\[D_{1} = 4 - 2 = 2\]

\[x_{1,2} = \frac{2 \pm \sqrt{2}}{2} = 1 \pm \frac{\sqrt{2}}{2}.\]

\[Ответ:x = 1 + \frac{\sqrt{2}}{2}.\]