Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 338

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\[1)\lg(x - 1) - \lg(2x - 11) = \lg 2\]

\[\lg\frac{x - 1}{2x - 11} = \lg 2\]

\[\frac{x - 1}{2x - 11} = 2\]

\[x - 1 = 2(2x - 11)\]

\[x - 1 = 4x - 22\]

\[- 3x = - 21\]

\[x = 7.\]

\[имеет\ смысл\ при:\]

\[x - 1 > 0\]

\[x > 1.\]

\[2x - 11 > 0\]

\[x > 5,5.\]

\[Ответ:\ \ x = 7.\]

\[2)\lg(3x - 1) - \lg(x + 5) = \lg 5\]

\[\lg\frac{3x - 1}{x + 5} = \lg 5\]

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

\[3x - 1 = 5(x + 5)\]

\[3x - 1 = 5x + 25\]

\[- 2x = 26\ \]

\[x = - 13.\]

\[имеет\ смысл\ при:\]

\[3x - 1 > 0\ \]

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

\[x + 5 > 0\]

\[\ x > - 5.\]

\[Ответ:\ \ нет\ решений.\]

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

\[\log_{3}\frac{x^{3} - x}{x} = \log_{3}3\]

\[\frac{x^{3} - x}{x} = 3\]

\[x^{3} - x = 3x\]

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

\[x\left( x^{2} - 4 \right) = 0\]

\[(x + 2) \bullet x \bullet (x - 2) = 0\]

\[x_{1} = - 2,\ \ \ x_{2} = 0,\ \ \ x_{3} = 2.\]

\[имеет\ смысл\ при:\]

\[x^{3} - x > 0\]

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

\[(x + 1) \bullet x \bullet (x - 1) > 0\]

\[- 1 < x < 0;\ x > 1.\]

\[имеет\ смысл\ при:\]

\[x > 0.\]

\[Ответ:\ \ x = 2.\]