Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 363

\[\boxed{\mathbf{363}\mathbf{.}}\]

\[\log_{0,2}x + \log_{0,2}(x - 2) < \log_{0,2}3\]

\[\log_{0,2}\left( x(x - 2) \right) < \log_{0,2}3\]

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

\[x^{2} - 2x > 3\]

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

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

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

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

\[(x + 2)(x - 3) > 0\]

\[x < - 2\ ;\text{\ \ }x > 3.\]

\[имеет\ смысл\ при:\]

\[x > 0;\]

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

\[Ответ:\ \ x > 3.\]

\[\lg x - \log_{10^{- 1}}(x - 1) > \log_{10^{- 1}}\frac{1}{2}\]

\[\lg x + \lg(x - 1) > - \lg\left( \frac{1}{2} \right)\]

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

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

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

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

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

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

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

\[(x + 1)(x - 2) > 0\]

\[x < - 1;\ x > 2.\]

\[имеет\ смысл\ при:\]

\[x > 0;\]

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

\[Ответ:\ \ x > 2.\]