Решебник по алгебре 9 класс Мерзляк Задание 776

 

\[\boxed{\mathbf{776\ (776).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\left\{ \begin{matrix} a_{3} + a_{5} + a_{13} = 33 \\ a_{15} - a_{8} - a_{10} = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} a_{1} + 2d + a_{1} + 4d + a_{1} + 12d = 33 \\ a_{1} + 14d - a_{1} - 7d - a_{1} - 9d = - 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 3a_{1} + 18d = 33\ \ \ \ \ \ \ \\ - a_{1} - 2d = - 1\ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} 3a_{1} + 18d = 33 \\ - 3a_{1} - 6d = - 3 \\ \end{matrix} \right.\ + \ \ \ \ \ \ \ \]

\[\ \left\{ \begin{matrix} 12d = 30 \\ a_{1} = 1 - 2d \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} d = 2,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = 1 - 2 \cdot 2,5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} d = 2,5 \\ a_{1} = - 4 \\ \end{matrix} \right.\ \]

\[S_{33} = \frac{2a_{1} + 32d}{2} \cdot 33 =\]

\[= \frac{- 8 + 32 \cdot 2,5}{2} \cdot 33 =\]

\[= \frac{- 8 + 80}{2} \cdot 33 =\]

\[= 36 \cdot 33 = 1188\]

\[Ответ:1188.\]

\[\boxed{\mathbf{776.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[3 < m < 4,\ \ - 3 < n < - 2\]

\[1) + \left| \begin{matrix} 6 < 2m < 8 \\ - 9 < 3n < - 6 \\ \end{matrix} \right.\ \]

\[\text{\ \ \ \ }\overline{- 3 < 2m + 3n < 2}\]

\[2) + \left| \begin{matrix} 0,6 < 0,2m < 0,8 \\ 2 < - n < 3\ \\ \end{matrix} \right.\ \]

\[\text{\ \ \ \ }\overline{\ 2,6 < 0,2m - n < 3,8}\]

\[3) + \left| \begin{matrix} - 20 < - 5m < - 15 \\ - 12 < 4n < - 8\ \ \\ \end{matrix} \right.\ \]

\[\text{\ \ \ \ }\overline{- 32 < - 5m + 4n < - 23}\]

\[4) - \frac{1}{2} < - \frac{1}{n} < - \frac{1}{3},\ \ \]

\[- \frac{3}{2} < \frac{m}{n} < - \frac{4}{3}\]

\[3 + \frac{4}{3} < m - \frac{m}{n} < 4 + \frac{2}{3}\]

\[4\frac{1}{3} < m - \frac{m}{n} < 5,5\]