Решебник по алгебре 9 класс Макарычев Задание 675

\[\boxed{\text{675}\text{\ (675)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Пусть\ a_{1},a_{2},a_{3} - углы\ данного\ треугольника.\]

\[a_{1} + a_{2} + a_{3} = 180{^\circ}.\]

\[\Longrightarrow a_{1} = x,\ \ a_{2} = x + d,\ \ a_{3} = x + 2d;\]

\[\Longrightarrow x + x + d + x + 2d = 3x + 3d = 180{^\circ} \Longrightarrow\]

\[\Longrightarrow a_{2} = x + d = 60{^\circ}.\]

\[\boxed{\text{675.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{2} \cdot a_{6} = a_{3} \cdot a_{5};\]

\[a_{2} \cdot a_{6} = a_{1}q \cdot a_{1}q^{5} = a_{1}^{2} \cdot q^{6},\]

\[a_{3} \cdot a_{5} = a_{1}q² \cdot a_{1}q^{4} = a_{1}^{2}{\cdot q}^{6},\]

\[a_{2} \cdot a_{6} = a_{3} \cdot a_{5} \Longrightarrow ч.т.д.\]

\[\textbf{б)}\ a_{n - 3} \cdot a_{n + 8} = a_{n} \cdot a_{n + 5},\]

\[\text{\ \ }где\ n > 3,\]

\[\ a_{n - 3} \cdot a_{n + 8} =\]

\[= a_{1}q^{n - 4} \cdot a_{1}q^{n + 7} = a_{1}^{2}q^{2n + 3},\]

\[a_{n} \cdot a_{n + 5} = a_{1}q^{n - 1} \cdot a_{1}q^{n + 4} =\]

\[= a_{1}^{2}q^{2n + 3},\]

\[a_{n - 3} \cdot a_{n + 8} = a_{n} \cdot a_{n + 5} \Longrightarrow ч.т.д.\]