Решебник по алгебре 9 класс Мерзляк Задание 799

 

\[\boxed{\mathbf{799\ (799).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[x_{1} = - 3;\ \ x_{11} = 12\]

\[\left\{ \begin{matrix} x_{1} + 10d = 12 \\ x_{1} = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\ \right.\ ( - )\text{\ \ }\]

\[\ \left\{ \begin{matrix} 10d = 15 \\ x_{1} = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} d = 1,5 \\ x_{1} = - 3 \\ \end{matrix} \right.\ \]

\[x_{10} = x_{11} - d = 12 - 1,5 = 10,5\]

\[x_{25} = x_{1} + 24d =\]

\[= - 3 + 24 \cdot 1,5 = 33\]

\[n = 25 - 10 + 1 = 16\]

\[S_{16} = \frac{x_{10} + x_{25}}{2} \cdot 16 =\]

\[= \frac{10,5 + 33}{2} \cdot 16 =\]

\[\text{=}43,5 \cdot 8 = 348.\]

\[Ответ:348.\]

\[\boxed{\mathbf{799.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[y = x^{2} + bx + 2\]

\[1)\ x² + bx + 2 = 0\ \]

\[D = b^{2} - 8\ \ \]

\[b^{2} - 8 = 0\]

\[b^{2} = 8 \Longrightarrow \ \ b = \pm 2\sqrt{2};\]

\[2)\ b² - 8 < 0\ \]

\[\ b^{2} < 8 \Longrightarrow \ \ b \in \left( - 2\sqrt{2};2\sqrt{2} \right);\]

\[3)\ \left\{ \begin{matrix} x_{1} + x_{2} = - b \\ x_{1}x_{2} = 2\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[x_{1} - x_{2} = 4;\ \ x_{1} = 4 + x_{2}\]

\[\left\{ \begin{matrix} 4 + x_{2} + x_{2} = - b \\ \left( 4 + x_{2} \right)x_{2} = 2\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 4 + 2x_{2} = - b \\ 4x_{2} + x_{2}^{2} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x_{2} = \frac{- b - 4}{2} \\ x_{1} = \frac{4 - b}{2}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{- b - 4}{2} \cdot \frac{4 - b}{2} = 2\ \ \ | \cdot 4\]

\[( - b - 4)(4 - b) = 8\]

\[- 4b + b^{2} - 16 + 4b = 8\]

\[b² = 24 \Longrightarrow \ \ b = \pm 2\sqrt{6}.\]