Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 913

\[1)\log_{0,5}\left( x^{2} - 5x + 6 \right) > - 1\]

\[\log_{\frac{1}{2}}\left( x^{2} - 5x + 6 \right) > \log_{\frac{1}{2}}\left( \frac{1}{2} \right)^{- 1}\]

\[x^{2} - 5x + 6 < 2\]

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

\[D = 25 - 16 = 9\]

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

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

\[(x - 1)(x - 4) < 0\]

\[1 < x < 4.\]

\[Область\ определения:\]

\[x^{2} - 5x + 6 > 0\]

\[D = 25 - 24 = 1\]

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

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

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

\[x < 2;\ \ \ x > 3.\]

\[Ответ:\ \ x \in (1;\ 2) \cup (3;\ 4).\]

\[2)\log_{8}\left( x^{2} - 4x + 3 \right) \leq 1\]

\[\log_{8}\left( x^{2} - 4x + 3 \right) \leq \log_{8}8^{1}\]

\[x^{2} - 4x + 3 \leq 8\]

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

\[D = 16 + 20 = 36\]

\[x_{1} = \frac{4 - 6}{2} = - 1;\]

\[x_{2} = \frac{4 + 6}{2} = 5;\]

\[(x + 1)(x - 5) \leq 0\]

\[- 1 \leq x \leq 5.\]

\[Область\ определения:\]

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

\[D = 16 - 12 = 4\]

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

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

\[(x - 1)(x - 3) > 0\]

\[x < 1;\ \text{\ \ }x > 3.\]

\[Ответ:\ \ x \in \lbrack - 1;\ 1) \cup (3;\ 5\rbrack.\]