Решебник по алгебре 9 класс Макарычев Задание 889

 

\[\boxed{\text{889\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{2} = - 6,\ \ a_{3} = - 2,\ \ \]

\[a_{15} = ?\]

\[\left\{ \begin{matrix} a_{2} = a_{1} + d\ \ \\ a_{3} = a_{1} + 2d \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} a_{1} + d = - 6\ \ \\ a_{1} + 2d = - 2 \\ \end{matrix} \right.\ \ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} a_{1} + d = - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ d = - 2 - ( - 6) = 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} d = 4\ \ \ \ \ \ \ \ \\ a_{1} = - 10, \\ \end{matrix} \right.\ \]

\[a_{15} = a_{1} + 14d =\]

\[= - 10 + 14 \cdot 4 = 46.\]

\[\textbf{б)}\ x_{2} = - 2,4;\ \ \ d = 1,2;\ \ \ S_{10} = ?\]

\[x_{2} = x_{1} + d\]

\[x_{1} = x_{2} - d = - 24 - 12 = - 36.\]

\[S_{10} = \frac{2x_{1} + d(n - 1)}{2} \cdot n =\]

\[= \frac{- 7,2 + 1,2 \cdot 9}{2} \cdot 10 = 18.\]

\[\boxed{\text{889.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[a^{2} + 4b^{2} - 5ab = 0\]

\[\left( a^{2} - ab \right) + \left( 4b^{2} - 4ab \right) = 0\]

\[a(a - b) - 4b(a - b) = 0\]

\[(a - 4b)(a - b) = 0.\]

\[1)\ a = b,\]

\[\sqrt[3]{65 + x} = \sqrt[3]{65 - x} \Longrightarrow x = 0;\]

\[2)\ a = 4b,\]

\[\sqrt[3]{65 + x} = 4\sqrt[3]{65 - x} \Longrightarrow x = 63.\]

\[Ответ:\ x = 0;x = 63.\]