\[\boxed{\text{277\ (277).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ y = \frac{7}{x + 5}\ убывает\ на\ \]
\[( - 5; + \infty),\ так\ как:\]
\[x_{1} = 2 \rightarrow \ \ \ \ y_{1} = 1\]
\[x_{2} = 9 \rightarrow \ \ y_{2} = 0,5\]
\[x_{2} > x_{1};\ \ y_{2} < y_{1}.\]
\[2)\ y = 6x - x^{2}\ возрастает\ на\]
\[\ ( - \infty;3\rbrack,\ так\ как:\]
\[x_{1} = 0, \rightarrow \ y_{1} = 0\]
\[x_{2} = 1 \rightarrow \ \ y_{2} = 6 - 1 = 5\]
\[x_{2} > x_{1};\ \ y_{2} > y_{1}\text{.\ }\]
\[\boxed{\text{277.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \left\{ \begin{matrix} - x < 2 \\ 2x > 7 \\ x < - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x > - 2 \\ x > 3,5 \\ x < - 4 \\ \end{matrix} \right.\ \ \]
\[Ответ:\ \varnothing.\ \]
\[2)\ \left\{ \begin{matrix} 3x - 1 < 2x + 2 \\ 2x + 1 > 8 - 5x \\ 5x - 25 \leq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x < 3\ \ \ \ \\ 7x > 7\ \ \\ 5x \leq 25 \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x < 3 \\ x > 1 \\ x \leq 5 \\ \end{matrix} \right.\ \]
\[Ответ:x \in (1;3)\text{.\ }\]