Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 395

\[\boxed{\mathbf{395}\mathbf{.}}\]

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

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

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

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

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

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[x_{1} = \frac{1 - 3}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{1 + 3}{2} = 2.\]

\[имеет\ смысл\ при:\]

\[x - 1 > 0 \Longrightarrow x > 1;\]

\[x > 0.\]

\[Ответ:\ \ x = 2.\]

\[2)\log_{\frac{1}{2}}{\frac{10}{7 - x} = \log_{\frac{1}{2}}x}\]

\[\frac{10}{7 - x} = x\]

\[10 = x(7 - x)\]

\[10 = 7x - x^{2}\]

\[x^{2} - 7x + 10 = 0\]

\[D = 7^{2} - 4 \bullet 10 = 49 - 40 = 9\]

\[x_{1} = \frac{7 - 3}{2} = 2;\text{\ \ }x_{2} = \frac{7 + 3}{2} = 5.\]

\[имеет\ смысл\ при:\]

\[7 - x > 0 \Longrightarrow x < 7;\]

\[x > 0.\]

\[Ответ:\ \ x_{1} = 2;\ \ x_{2} = 5.\]

\[3)\lg\frac{x + 8}{x - 1} = \lg x\]

\[\frac{x + 8}{x - 1} = x\]

\[x + 8 = x(x - 1)\]

\[x + 8 = x^{2} - x\]

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

\[D = 2^{2} + 4 \bullet 8 = 4 + 32 = 36\]

\[x_{1} = \frac{2 - 6}{2} = - 2;\text{\ \ }\]

\[x_{2} = \frac{2 + 6}{2} = 4\]

\[имеет\ смысл\ при:\]

\[1)\ \frac{x + 8}{x - 1} > 0\]

\[(x + 8)(x - 1) > 0\]

\[x < - 8\ \ и\ \ x > 1.\]

\[2)\ x > 0.\]

\[Ответ:\ \ x = 4.\]

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

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

\[x - 4 = x(x - 2)\]

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

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

\[D = 3^{2} + 4 \bullet 4 = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1;\text{\ \ }\]

\[x_{2} = \frac{3 + 5}{2} = 4.\]

\[имеет\ смысл\ при:\]

\[1)\ \frac{x - 4}{x - 2} > 0\]

\[(x - 2)(x - 4) > 0\]

\[x < 2\ \ и\ \ x > 4.\]

\[2)\ x > 0.\]

\[Ответ:\ \ решений\ нет.\]