Решебник по алгебре 9 класс Мерзляк Задание 800

 

\[\boxed{\mathbf{800\ (800).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S_{6} = 39;\ \ S_{14} = - 77\]

\[S_{6} = \frac{2a_{1} + 5d}{2} \cdot 6 =\]

\[= \left( 2a_{1} + 5d \right) \cdot 3\]

\[S_{14} = \frac{2a_{1} + 13d}{2} \cdot 14 =\]

\[= \left( 2a_{1} + 13d \right) \cdot 7\]

\[\left\{ \begin{matrix} \left( 2a_{1} + 5d \right) \cdot 3 = 39\ \ \ \ \ \ \ \ \ |\ :3 \\ \left( 2a_{1} + 13d \right) \cdot 7 = - 77\ \ \ |\ :7 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ \ \ \ }\left\{ \begin{matrix} 2a_{1} + 5d = 13\ \ \ \ \ \ \\ 2a_{1} + 13d = - 11 \\ \end{matrix} \right.\ \ ( - )\]

\[\text{\ \ }\left\{ \begin{matrix} - 8d = 24\ \ \ \ \ \ \ \\ a_{1} = \frac{13 - 5d}{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} d = - 3\ \ \ \ \ \ \ \ \ \ \ \\ a_{1} = \frac{13 + 15}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} d = - 3 \\ a_{1} = 14 \\ \end{matrix} \right.\ \]

\[Ответ:\ a_{1} = 14;\ \ d = - 3.\]

\[\boxed{\mathbf{800.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[y = x^{2} + px + q;\ \ A\ (1;1);\]

\[\text{\ \ }\text{B\ }(2;2)\]

\[\left\{ \begin{matrix} 1 + p + q = 1 \\ 4 + 2p + q = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} p + q = 0\ \ \ \ \ \ \\ 2p + q = - 2 \\ \end{matrix} \right.\ - \ \ \ \ \left\{ \begin{matrix} - p = 2 \\ q = - p \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} p = - 2 \\ q = 2\ \ \ \ \\ \end{matrix} \right.\ \]

\[Функция\ имеет\ вид:\ \ \]

\[y = x^{2} - 2x + 2.\]

\[1)\ C\ ( - 1; - 1):\ \ \]

\[- 1 = ( - 1)^{2} - 2 \cdot ( - 1) + 2\ \ \]

\[- 1 = 1 + 2 + 2 \Longrightarrow \ \ - 1 \neq 5\]

\[не\ проходит.\]

\[2)\ \text{D\ }(3;5):\ \]

\[5 = 3^{2} - 2 \cdot 3 + 2\]

\[\ 5 = 9 - 6 + 2\ \ \]

\[5 = 5\]

\[проходит.\]