ГДЗ по алгебре 11 класс Никольский Параграф 10. Равносильность уравнений на множествах Задание 41

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\[5x - 1 > 0\]

\[x > 0,2;\]

\[5x^{2} - x > 0\]

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

\[x < 0;x > 0,2.\]

\[3x > 0\]

\[x > 0.\]

\[3x \neq 1\]

\[x \neq \frac{1}{3}.\]

\[M = \left( 0,2;\frac{1}{3} \right) \cup \left( \frac{1}{3}; + \infty \right).\]

\[\log_{3}\left( 5x^{2} - x \right) =\]

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

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

\[\log_{3}x = a;\ \ \log_{3}(5x - 1) = b:\]

\[ab + b^{2} - 1 - a = 0\]

\[(b - 1)(a + b + 1) = 0\]

\[b - 1 = 0;\]

\[a + b + 1 = 0.\]

\[\log_{3}(5x - 1) - 1 = 0\]

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

\[5x - 1 = 3\]

\[5x = 4\]

\[x = 0,8.\]

\[\log_{3}x + \log_{3}(5x - 1) + 1 = 0\]

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

\[5x^{2} - x = \frac{1}{3}\]

\[15x^{2} - 3x - 1 = 0\]

\[D = 9 + 60 = 69\]

\[x_{1} = \frac{3 + \sqrt{69}}{30};\]

\[x_{2} = \frac{3 - \sqrt{69}}{30} < 0.\]

\[Ответ:x = 0,8;\ \ x = \frac{3 + \sqrt{69}}{30}.\]

\[20x - 1 > 0\]

\[20x > 1\]

\[x > 0,05.\]

\[20x^{2} - x > 0\]

\[20x(x - 0,05) > 0\]

\[x < 0;\ \ x > 0,05.\]

\[16x > 0\]

\[x > 0.\]

\[16x \neq 1\]

\[x \neq \frac{1}{16}.\]

\[M = \left( 0,05;\frac{1}{16} \right) \cup \left( \frac{1}{16}; + \infty \right).\]

\[1)\ \log_{4}(20x - 1) - 2 = 0\]

\[\log_{4}(20x - 1) = 2\]

\[20x - 1 = 2^{4}\]

\[20x - 1 = 16\]

\[20x = 17\]

\[x = \frac{17}{20}\]

\[x = 0,85.\]

\[2)\log_{4}x + \log_{4}(20x - 1) + 2 =\]

\[= 0\]

\[20x^{2} - x = 4^{- 2}\]

\[2x^{2} - x = \frac{1}{16}\]

\[320x^{2} - 16x - 1 = 0\]

\[D_{1} = 64 + 320 = 384\]

\[x_{1} = \frac{8 + 8\sqrt{6}}{320} = \frac{1 + \sqrt{6}}{40};\]

\[x_{2} = \frac{8 - 8\sqrt{6}}{320} = \frac{1 - \sqrt{6}}{40} < 0.\]

\[Ответ:x = 0,85;\ \ x = \frac{1 + \sqrt{6}}{40}.\]