Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 909

\[\boxed{\mathbf{909}.}\]

\[1)\log_{3 - x}\frac{4 - x}{5 - x} \leq 1\]

\[Область\ определения:\]

\[\frac{4 - x}{5 - x} > 0\]

\[x < 4;\ \ x > 5.\]

\[Если\ 0 < 3 - x < 1;\]

\[то\ 2 < x < 3:\]

\[\left\{ \begin{matrix} 2 < x < 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{4 - x}{5 - x} - (3 - x) \geq 0 \\ \end{matrix} \right.\ \]

\[- \left( x - \frac{7 - \sqrt{5}}{2} \right)\left( x - \frac{7 + \sqrt{5}}{2} \right) \geq 0\]

\[\left( x - \frac{7 - \sqrt{5}}{2} \right)\left( x - \frac{7 + \sqrt{5}}{2} \right) \leq 0\]

\[\frac{7 - \sqrt{5}}{2} \leq x \leq \frac{7 + \sqrt{5}}{2}\]

\[\frac{7 - \sqrt{5}}{2} \leq x < 3.\]

\[Если\ 3 - x > 1;то\ x < 2:\]

\[\left\{ \begin{matrix} x < 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{4 - x}{5 - x} - (3 - x) \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[- \left( x - \frac{7 - \sqrt{5}}{2} \right)\left( x - \frac{7 + \sqrt{5}}{2} \right) \leq 0\]

\[\left( x - \frac{7 - \sqrt{5}}{2} \right)\left( x - \frac{7 + \sqrt{5}}{2} \right) \geq 0\]

\[x \leq \frac{7 - \sqrt{5}}{2};\ \ \ x \geq \frac{7 + \sqrt{5}}{2}.\]

\[x < 2.\]

\[Найдем\ корни\ уравнения:\]

\[\frac{4 - x}{5 - x} - (3 - x) = 0\]

\[4 - x - (3 - x)(5 - x) = 0\]

\[4 - x - 15 + 5x + 3x - x^{2} = 0\]

\[- x^{2} + 7x - 11 = 0\]

\[x^{2} - 7x + 11 = 0\]

\[D = 49 - 44 = 5\]

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

\[Ответ:\]

\[x \in ( - \infty;2) \cup \left\lbrack \frac{7 - \sqrt{5}}{2};3 \right).\]

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

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

\[\log\left( x^{2} - 5x + 6 \right) = \log{2x}\]

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

\[x^{2} - 7x + 6 = 0\]

\[(x - 6)(x - 1) < 0.\]

\[Область\ определения:\]

\[0 < x < \frac{1}{2};\]

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

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

\[Ответ:0 < x < \frac{1}{2};\ \ 1 < x < 2;\ \]

\[\ 3 < x < 6.\]