Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 20

\[\boxed{\mathbf{20.}}\]

\[- x^{2} + 6x - 8 > 0\]

\[x^{2} - 6x + 8 < 0\]

\[D_{1} = 9 - 8 = 1\]

\[x_{1} = 3 + 1 = 4;\]

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

\[2 < x < 4.\]

\[M = (2;4).\]

\[\log_{0,2}\left( - x^{2} + 6x - 8 \right) > 0\]

\[\log_{0,2}\left( - x^{2} + 6x - 8 \right) > \log_{0,2}1\]

\[- x^{2} + 6x - 8 < 1\]

\[x^{2} - 6x + 9 > 0\]

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

\[x \neq 3.\]

\[M_{1} = (2;3) \cup (3;4) - не\ имеет\ \]

\[решений.\]

\[При\ x = 3:\]

\[\log_{0,2}1 \leq 0\]

\[0 \leq 0.\]

\[Ответ:x = 3.\]

\[0 \leq cos^{2}x \leq 1\]

\[0 \leq 3cos^{2}x \leq 3.\]

\[\left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| \geq 0\]

\[3 + \left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| \geq 3.\]

\[Равносильная\ система\ \]

\[уравнений:\]

\[\left\{ \begin{matrix} 3 + \left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| = 3 \\ 3cos^{2}x = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| = 0 \\ \text{co}s^{2}x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| = 0\]

\[\left| \log_{5}\left( x^{2} - 4x + 1 \right) \right| = \log_{5}1\]

\[x^{2} - 4x + 1 = 1\]

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

\[x(x - 4) = 0\]

\[x = 0;\ \ x = 4.\]

\[Проверим:\]

\[\text{co}s^{2}x = cos^{2}0 = 1.\]

\[\text{co}s^{2}x = cos^{2}4 \neq 1.\]

\[Ответ:x = 0.\]