Решебник по алгебре 9 класс Мерзляк Задание 1025

\[\boxed{\mathbf{1025\ (1025).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1) - a_{2},\ - a_{4},\ - a_{6},\ - a_{8},\ldots\]

\[- \left( a_{1} + d \right),\ - \left( a_{1} + 3d \right),\ \]

\[- \left( a_{1} + 5d \right),\ - \left( a_{1} + 7d \right)\ldots\]

\[- \left( a_{1} + 3d \right) - \left( - \left( a_{1} + d \right) \right) =\]

\[= - a_{1} - 3d + a_{1} + d =\]

\[= - 2d \Longrightarrow \ \ является.\]

\[2)\ a_{1} + 5,\ a_{2} + 5,\ a_{3} + 5,\ \ldots;\]

\[\ a_{1} + 5,\ \ a_{1} + d + 5,\]

\[\ \ a_{1} + 2d + 5,\ \ldots\]

\[\ a_{1} + d + 5 - \ a_{1} - 5 =\]

\[= d \Longrightarrow \ \ является.\]

\[3)\ 1 - a_{1},\ 1 - a_{2},\ 1 - a_{3},\ldots\]

\[1 - a_{1},1 - a_{1} - d,\]

\[\ 1 - a_{1} - 2d,\ldots\]

\[1 - a_{1} - d - 1 + a_{1} =\]

\[= - d \Longrightarrow \ является.\]

\[4)\ a_{1}²,\ a_{2}²,\ a_{3}²,\ldots\]

\[a_{1}²,\ \left( a_{1} + d \right)^{2},\ \ \ \left( a_{1} + 2d \right)^{2},\ldots\]

\[\left( a_{1} + d \right)^{2} - a_{1}^{2} = a_{1}^{2} + 2a_{1}d +\]

\[+ d^{2} - a_{1}^{2} = 2a_{1}d + d^{2}\text{\ \ \ \ }или\ \ \]

\[\left( a_{1} + 2d \right)^{2} - \left( a_{1} + d \right)^{2} = a_{1}^{2} +\]

\[+ 4a_{1}d + 4d^{2} - a_{1}^{2} - 2a_{1}d -\]

\[- d^{2} = 2a_{1}d + 3d^{2}\]

\[Разности\ не\ равны,\ не\ \]

\[является.\]

\[5)\ a_{1} + a_{2},\ a_{2} + a_{3},\ a_{3} + a_{4},\ldots\]

\[a_{1} + a_{1} + d,\ a_{1} + d + a_{1} +\]

\[+ 2d,\ a_{1} + 2d + a_{1} + 3d,\ldots\]

\[2a_{1} + d,\ 2a_{1} + 3d,\ 2a_{1} + 5d,\ldots\]

\[2a_{1} + 3d - 2a_{1} - d = 2d\ \ \ \ или\ \ \ \ \]

\[\ 2a_{1} + 5d - 2a_{1} - 3d = 2d\]

\[Является.\]

\[6)\ a_{1} + a_{2},\ a_{3} + a_{4},\ a_{5} + a_{6},\ldots\]

\[a_{1} + a_{1} + d,\ a_{1} + 2d + a_{1} +\]

\[+ 3d,\ a_{1} + 4d + a_{1} + 5d,\ldots\]

\[2a_{1} + d,\ 2a_{1} + 5d,\ 2a_{1} + 9d,\ldots\]

\[2a_{1} + 5d - 2a_{1} - d = 4d\ \ \ или\ \ \ \ \ \]

\[\ 2a_{1} + 9d - 2a_{1} - 5d = 4d\]

\[Является.\]