Решебник по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 40

\[\boxed{\mathbf{40.}}\]

\[4x > 0\]

\[x > 0.\]

\[4x \neq 1\]

\[x \neq \frac{1}{4}.\]

\[8x^{2} - x > 0\]

\[8x\left( x - \frac{1}{8} \right) > 0\]

\[x < 0;\ \ x > \frac{1}{8}.\]

\[8x - 1 > 0\]

\[x > \frac{1}{8}.\]

\[M = \left( \frac{1}{8};\frac{1}{4} \right) \cup \left( \frac{1}{4}; + \infty \right).\]

\[\log_{2}\left( 8x^{2} - x \right) =\]

\[= \log_{2}{x(8x - 1)} =\]

\[= \log_{2}x + \log_{2}(8x - 1);\]

\[\log_{4x}(8x - 1) = \frac{\log_{2}(8x - 1)}{\log_{2}{4x}} =\]

\[= \frac{\log_{2}(8x - 1)}{2 + \log_{2}x};\]

\[1)\ 2x - \frac{1}{4} = 1\]

\[2x = \frac{5}{4}\]

\[x = \frac{5}{8}.\]

\[2)\ 32x^{2} - 4x = 1\]

\[32x^{2} - 4x - 1 = 0\]

\[D_{1} = 4 + 32 = 36\]

\[x_{1} = \frac{2 + 6}{32} = \frac{8}{32} =\]

\[= \frac{1}{4}\ (не\ подходит).\]

\[x_{2} = \frac{2 - 6}{32} =\]

\[= - \frac{1}{8}\ (не\ подходит).\]

\[Ответ:x = \frac{5}{8}.\]

\[\textbf{б)}\log_{5}{(x - 2)} \cdot \log_{\sqrt{x + 10}}5 = 1\]

\[x - 2 > 0\]

\[x > 2.\]

\[x + 10 > 0\]

\[x > - 10.\]

\[x + 10 \neq 1\]

\[x \neq - 9.\]

\[M = (2; + \infty).\]

\[\log_{\sqrt{x + 10}}5 = \frac{1}{\log_{5}\sqrt{x + 10}} =\]

\[= \frac{1}{\frac{1}{2}\log_{5}(x + 10)};\]

\[\frac{2\log_{5}(x - 2)^{2}}{\log_{5}{(x + 10)}} = 1\]

\[\log_{5}(x - 2)^{2} = \log_{5}{(x + 10)}\]

\[(x - 2)^{2} = x + 10\]

\[x^{2} - 4x + 4 - x - 10 = 0\]

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

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

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

\[Ответ:x = 6.\]

\[\textbf{в)}\log_{3}{(x - 1)} \cdot \log_{\sqrt{x + 5}}3 = 1\]

\[x - 1 > 0\]

\[x > 1.\]

\[x + 5 > 0\]

\[x > - 5.\]

\[x + 5 \neq 1\]

\[x \neq - 4.\]

\[M = (1; + \infty).\]

\[\log_{\sqrt{x + 5}}3 = \frac{1}{\log_{3}\sqrt{x + 5}} =\]

\[= \frac{1}{\frac{1}{2}\log_{3}{(x + 5)}} = \frac{2}{\log_{3}{(x + 5)}};\]

\[\log_{3}(x - 1) \cdot \frac{2}{\log_{3}(x + 5)} = 1\]

\[\frac{2\log_{3}(x - 1)}{\log_{3}(x + 5)} = 1\]

\[\log_{3}(x - 1)^{2} = \log_{3}(x + 5)\]

\[(x - 1)^{2} = x + 5\]

\[x^{2} - 2x + 1 - x - 5 = 0\]

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

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

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

\[Ответ:x = 4.\]