Решебник по алгебре 9 класс Макарычев Задание 697

 

\[\boxed{\text{697}\text{\ (697)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ S_{20} = 1000,\ \ S_{40} = 10000;\]

\[S_{20} = \frac{2a_{1} + 19d}{2} \cdot 20 = 10 \cdot \left( 2a_{1} + 19d \right) = 1000 \Longrightarrow\]

\[2a_{1} + 19d = 100;\]

\[S_{40} = \frac{2a_{1} + 39d}{2} \cdot 40 = 20 \cdot \left( 2a_{1} + 39d \right) = 1000 \Longrightarrow\]

\[2a_{1} + 39d = 500;\]

\[\left\{ \begin{matrix} 2a_{1} + 19d = 100 \\ 2a_{1} + 39d = 500 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 20d = 400\ \ \ \ \ \ \ \\ a_{1} = \frac{100 - 19d}{2} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = 20\ \ \ \ \ \ \ \ \\ a_{1} = - 140 \\ \end{matrix} \right.\ \Longrightarrow\]

\[a_{50} = a_{1} + d(n - 1) = - 140 + 20 \cdot 49 = 840;\]

\[\textbf{б)}\ S_{5} = 0,5;\ \ \ \ \ S_{15} = - 81;\]

\[S_{5} = \frac{2a_{1} + 4d}{2} \cdot 5 = 5 \cdot \left( a_{1} + 2d \right) = 0,5 \Longrightarrow a_{1} + 2d = 0,1;\]

\[S_{15} = \frac{2a_{1} + 14d}{2} \cdot 15 = 15 \cdot \left( a_{1} + 7d \right) = - 81 \Longrightarrow a_{1} + 7d = - 5,4;\]

\[\left\{ \begin{matrix} a_{1} + 2d = 0,1\ \ \\ a_{1} + 7d = - 5,4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 5d = - 5,5\ \ \ \ \ \\ a_{1} = 0,1 - 2d \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = - 1,1 \\ a_{1} = 2,3. \\ \end{matrix} \right.\ \]

\[a_{50} = a_{1} + 49d = 2,3 - 1,1 \cdot 49 = - 51,6.\]

\[\boxed{\text{697.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[0,45 \approx 0,5\]

\[|0,5 - 0,45| = 0,05;\]

\[\frac{0,05}{0,45} = \frac{1}{9} = 0,(1) \approx 10\%.\]

\[2,53 \approx 2,5\]

\[|2,5 - 2,53| = 0,03;\]

\[\frac{0,03}{2,53} = \frac{3}{253} =\]

\[= 0,01185\ldots \approx 1,2\%.\]

\[31,98 \approx 32,0\]

\[|32,0 - 31,98| \approx 0,02\]

\[\frac{0,02}{31,98} = \frac{2}{3198} =\]

\[= 0,000625\ldots \approx 0,0625\%.\]