Решебник по алгебре 9 класс Мерзляк Задание 862

 

\[\boxed{\mathbf{862\ (862).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Арифметическая\ прогрессия:\]

\[\ a_{1} + a_{2} + a_{3} = 30.\ \ \]

\[a_{1} + a_{1} + d + a_{1} + 2d = 30\ \]

\[\ 3a_{1} + 3d = 30\ \ |\ :3\ \]

\[a_{1} + d = 10\]

\[d = 10 - a_{1}.\]

\[Геометрическая\ прогрессия:\ \]

\[a_{1} - 5,\ a_{2} - 4,\ a_{3}\text{.\ }\]

\[\left( a_{2} - 4 \right)^{2} = a_{3}\left( a_{1} - 5 \right)\text{\ \ \ }\]

\[\left( a_{1} + d - 4 \right)^{2} =\]

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

\[(10 - 4)^{2} =\]

\[= \left( a_{1} + 2 \cdot \left( 10 - a_{1} \right) \right) \cdot \left( a_{1} - 5 \right)\]

\[6^{2} = \left( a_{1} + 20 - 2a_{1} \right)\left( a_{1} - 5 \right)\]

\[36 = \left( - a_{1} + 20 \right)\left( a_{1} - 5 \right)\]

\[36 = - a_{1}^{2} + 5a_{1} + 20a_{1} - 100\]

\[a_{1}^{2} - 25a_{1} + 136 = 0\]

\[a_{1} + a_{2} = 25,\ \ a_{1} = 8\]

\[a_{1}a_{2} = 136,\ \ a_{1} = 17.\]

\[при\ a_{1} = 8 \Longrightarrow \ d = 10 - 8 = 2:\ \ \]

\[a_{2} = 8 + 2 = 10,\ \ \]

\[a_{3} = 10 + 2 = 12;\]

\[при\ a_{1} = 17 \Longrightarrow \ \ d =\]

\[= 10 - 17 = - 7:\ \ \]

\[a_{2} = 17 - 7 = 10,\ \ \]

\[a_{3} = 10 - 7 = 3.\]

\[Ответ:8,\ 10,\ 12\ \ \ или\ \ 17,\ 10,\ 3.\]

\[\boxed{\mathbf{862.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ b_{2},\ b_{4},\ b_{6},\ldots = b_{1}q,\ b_{1}q^{3},\ b_{1}q^{5},\ \ \]

\[является \Longrightarrow \frac{b_{1}q^{3}}{b_{1}q} = q^{2};\]

\[2)\ b_{1} + 1,\ b_{2} + 1,\ b_{3} + 1,\ldots =\]

\[= b_{1}^{2},\ b_{1}^{2}q^{2},\ b_{1}^{2}q^{4},\ \ \]

\[является \Longrightarrow \ \ \ \frac{b_{1}^{2}q^{2}}{b_{1}^{2}} = q^{2};\]

\[4) - b_{1},\ - b_{3},\ - b_{5},\ldots =\]

\[= - b_{1},\ - b_{1}q^{2}, - b_{1}q^{4},\ \]

\[\ является \Longrightarrow \ \ \frac{- b_{1}q^{2}}{- b_{1}} = q^{2};\]

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

\[= b_{1} + b_{1}q,\ b_{1}q + b_{1}q^{2},\ b_{1}q^{2} +\]

\[+ b_{1}q^{3},\ldots\]

\[является \Longrightarrow \text{\ \ \ }\frac{b_{1}q + b_{1}q^{2}}{b_{1} + b_{1}q} =\]

\[= \frac{b_{1}q(1 + q)}{b_{1}(1 + q)} = q;\]

\[6)\frac{1}{b_{1}},\frac{1}{b_{2}},\frac{1}{b_{3}},\ldots =\]

\[= \frac{1}{b_{1}},\frac{1}{b_{1}q},\frac{1}{b_{1}q^{2}},\ldots\]

\[является\ \Longrightarrow \frac{1}{b_{1}q}\ :\frac{1}{b_{1}} =\]

\[= \frac{b_{1}}{b_{1}q} = \frac{1}{q}.\]