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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 854
Задание 854
\[\boxed{\mathbf{854}.}\]
\[1)\ \lg^{2}\ (x + 1) =\]
\[= \lg(x + 1)\lg(x - 1) +\]
\[+ 2\lg^{2}{(x - 1)}\ \]
\[ОДЗ:\]
\[x + 1 > 0;\ \ \ \ \ \ x - 1 > 0\]
\[x > - 1;\ \ \ \ \ \ \ \ \ \ x > 1\]
\[x > 1.\]
\[1)\ \lg^{2}\ (x - 1) = 0\]
\[x - 1 = 10^{0}\]
\[x = 2.\]
\[Подставим\ в\ уравнение:\]
\[\lg^{2}3 = \lg 3 \cdot \lg 2 + 2\lg^{2}1\]
\[\lg^{2}3 \neq \lg 3 \cdot \lg 2 + 0\]
\[x = 2\ не\ является\ корнем\ \]
\[уравнения.\]
\[2)\ \frac{\ \lg^{2}(x + 1)}{\lg^{2}(x - 1)} =\]
\[= \frac{\lg(x + 1)\lg(x - 1)}{\lg^{2}(x - 1)} +\]
\[+ \frac{2\lg^{2}(x - 1)}{\lg^{2}(x - 1)}\ \]
\[\left( \frac{\lg(x + 1)}{\lg(x - 1)} \right)^{2} = \frac{\lg(x + 1)}{\lg(x - 1)} + 2\]
\[t = \frac{\lg(x + 1)}{\lg(x - 1)}:\]
\[t^{2} - t - 2 = 0\]
\[t_{1} + t_{2} = 1;\ \ \ t_{1} \cdot t_{2} = - 2\]
\[t_{1} = 2;\ \ \ t_{2} = - 1.\]
\[\textbf{а)}\ \frac{\lg(x + 1)}{\lg(x - 1)} = 2\]
\[\lg(x + 1) = 2\lg(x - 1)\]
\[\lg(x + 1) = \lg(x - 1)^{2}\]
\[x + 1 = x^{2} - 2x + 1\]
\[x^{2} - 3x = 0\]
\[x(x - 3) = 0\]
\[x = 0\ (не\ подходит\ по\ ОДЗ);\]
\[x = 3.\]
\[\textbf{б)}\ \frac{\lg(x + 1)}{\lg(x - 1)} = - 1\]
\[\lg(x + 1) = - 1\lg(x - 1)\]
\[\lg(x + 1) = \lg(x - 1)^{- 1}\]
\[x + 1 = \frac{1}{x - 1}\]
\[(x + 1)(x - 1) = 1\]
\[x^{2} - 1 = 1\]
\[x^{2} = 2\]
\[x = - \sqrt{2}\ (не\ подходит\ по\ ОДЗ);\]
\[x = \sqrt{2}.\]
\[Ответ:\ \ x = \sqrt{2};x = 3.\]
\[2)2\log_{5}(4 - x) \cdot \log_{2x}(4 - x) =\]
\[= 3\log_{5}{(4 - x)} - \log_{5}{2x}\ \]
\[2\log_{5}^{2}(4 - x) \cdot \frac{\log_{5}(4 - x)}{\log_{5}{2x}} =\]
\[= 3\log_{5}(4 - x) - \log_{5}{2x}\]
\[\frac{2\log_{5}^{2}(4 - x)}{\log_{5}{2x}} = 3\log_{5}(4 - x) -\]
\[- \log_{5}{2x}\ \ \ \ \ \ \ \ \ \ \ | \cdot \log_{5}{2x}\]
\[2\left( \frac{\log_{5}(4 - x)}{\log_{5}{2x}} \right)^{2} =\]
\[= \frac{3\log_{5}{(4 - x)}}{\log_{5}{2x}} - 1\]
\[t = \frac{\log_{5}(4 - x)}{\log_{5}{2x}}:\]
\[2t^{2} - 3t + 1 = 0\]
\[D = 9 - 8 = 1\]
\[t_{1} = \frac{3 + 1}{4} = 1;\ \ \ \]
\[\ t_{2} = \frac{3 - 1}{4} = \frac{1}{2}.\]
\[\textbf{а)}\ \frac{\log_{5}(4 - x)}{\log_{5}{2x}} = \frac{1}{2}\]
\[\frac{2\log_{5}(4 - x) - \log_{2}{2x}}{\log_{5}{2x}} = 0\]
\[\log_{5}{(4 - x)^{2} = \log_{5}{2x}}\]
\[(4 - x)^{2} = 2x\]
\[16 - 8x + x^{2} - 2x = 0\]
\[x^{2} - 10x + 16 = 0\]
\[D_{1} = 25 - 16 = 9\]
\[x_{1} = 5 + 3 = 8;\ \ \]
\[\ x_{2} = 5 - 3 = 2.\]
\[\textbf{б)}\ \frac{\log_{5}(4 - x)}{\log_{5}{2x}} = 1\]
\[\frac{\log_{5}{(4 - x)}}{\log_{5}{(2x)}} = \frac{\log_{5}{2x}}{\log_{5}{2x}}\]
\[\log_{5}(4 - x) = \log_{5}{2x}\]
\[4 - x = 2x\]
\[3x = 4\]
\[x = \frac{4}{3}.\]
\[Проверка\ показывает,\ что\ x =\]
\[= 8\ не\ является\ корнем\]
\[\ уравнения.\]
\[Ответ:x = 2;\ \ x = \frac{4}{3}.\]
