Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1351

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\[1)\ln\frac{2}{x + 1} = \ln(x + 2)\]

\[\frac{2}{x + 1} = x + 2\ \ \ \ \ | \bullet (x + 1)\]

\[2 = (x + 1)(x + 2)\]

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

\[x^{2} + 3x = 0\]

\[x \bullet (x + 3) = 0\]

\[x_{1} = - 3\ \ и\ \ x_{2} = 0\]

\[Имеет\ смысл:\]

\[x + 1 > 0\]

\[x > - 1.\]

\[x + 2 > 0\]

\[x > - 2.\]

\[Ответ:\ \ x = 0.\]

\[2)\log_{3}{\sqrt{3x - 6} - \log_{3}\sqrt{x - 3}} = 1\]

\[\log_{3}\sqrt{\frac{3x - 6}{x - 3}} = \log_{3}3\]

\[\sqrt{\frac{3x - 6}{x - 3}} = 3\ \ \ \ \ | \bullet \sqrt{x - 3}\]

\[\sqrt{3x - 6} = 3\sqrt{x - 3}\]

\[3x - 6 = 9(x - 3)\]

\[3x - 6 = 9x - 27\]

\[- 6x = - 21\]

\[x = 3,5.\]

\[Имеет\ смысл:\]

\[3x - 6 > 0\]

\[x > 2.\]

\[x - 3 > 0\]

\[x > 3.\]

\[Ответ:\ \ x = 3,5.\]