Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 890

\[\boxed{\mathbf{890}.}\]

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

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

\[(x - 2)(x - 3) = 2\]

\[x^{2} - 3x - 2x + 6 = 2\]

\[x^{2} - 5x + 4 = 0\]

\[D = 5^{2} - 4 \bullet 4 = 25 - 16 = 9\]

\[x_{1} = \frac{5 - 3}{2} = 1;\text{\ \ }x_{2} = \frac{5 + 3}{2} = 4.\]

\[имеет\ смысл\ при:\]

\[x - 2 > 0 \Longrightarrow x > 2;\]

\[x - 3 > 0 \Longrightarrow x > 3.\]

\[Ответ:\ \ x = 4.\]

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

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

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

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

\[- 5 - 5x + x + x^{2} = 27\]

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

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

\[= 16 + 128 = 144\]

\[x_{1} = \frac{4 - 12}{2} = - 4;\text{\ \ }\]

\[x_{2} = \frac{4 + 12}{2} = 8.\]

\[имеет\ смысл\ при:\]

\[5 - x > 0 \Longrightarrow x < 5;\]

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

\[Ответ:\ \ x = - 4.\]

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

\[\lg\left( (x - 2) \bullet x \right) = \lg 3\]

\[x(x - 2) = 3\]

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

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[x_{1} = \frac{2 - 4}{2} = - 1;\text{\ \ }\]

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

\[имеет\ смысл\ при:\]

\[x - 2 > 0 \Longrightarrow x > 2;\]

\[x > 0.\]

\[Ответ:\ \ x = 3.\]

\[4)\log_{\sqrt{6}}(x - 1) +\]

\[+ \log_{\sqrt{6}}(x + 4) = \log_{\sqrt{6}}6\]

\[\log_{\sqrt{6}}\left( (x - 1)(x + 4) \right) = \log_{\sqrt{6}}6\]

\[(x - 1)(x + 4) = 6\]

\[x^{2} + 4x - x - 4 = 6\]

\[x^{2} + 3x - 10 = 0\]

\[D = 3^{2} + 4 \bullet 10 = 9 + 40 = 49\]

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

\[x_{2} = \frac{- 3 + 7}{2} = 2.\]

\[имеет\ смысл\ при:\]

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

\[x + 4 > 0 \Longrightarrow x > - 4.\]

\[Ответ:\ \ x = 2.\]