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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 864
Задание 864
\[\boxed{\mathbf{864\ (864).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[Геометрическая\ прогрессия:\]
\[b_{1},\ b_{2},\ b_{3};\ \ \ \ \ \ \ \]
\[\ b_{1} + b_{2} + \ b_{3} = 26\]
\[b_{1} + b_{1}q + b_{1}q^{2} = 26\ \ \]
\[b_{1}\left( 1 + q + q^{2} \right) = 26.\]
\[Арифметическая\ прогрессия:\]
\[\text{\ \ }b_{1} + 1,b_{2} + 6,\ b_{3} + 3\]
\[b_{2} + 6 = \frac{b_{1} + 1 + b_{3} + 3}{2}\ \]
\[\ 2 \cdot \left( b_{2} + 6 \right) = b_{1} + b_{3} + 4\]
\[2b_{2} + 12 = b_{1} + b_{3} + 4\]
\[\ b_{1} + b_{3} - 2b_{2} = 8\]
\[b_{1} + b_{1}q^{2} - 2b_{1}q = 8\ \ \]
\[b_{1}\left( 1 + q^{2} - 2q \right) = 8\]
\[Запишем\ систему:\]
\[\left\{ \begin{matrix} b_{1}\left( 1 + q + q^{2} \right) = 26 \\ b_{1}\left( 1 + q^{2} - 2q \right) = 8 \\ \end{matrix}\ \ |(\ :) \right.\ \]
\[\ \frac{b_{1}\left( 1 + q + q^{2} \right)}{b_{1}\left( 1 + q^{2} - 2q \right)} = \frac{26}{8}\]
\[\frac{1 + q + q^{2}}{1 + q^{2} - 2q} = \frac{13}{4}\]
\[4 \cdot \left( 1 + q + q^{2} \right) =\]
\[= 13 \cdot \left( 1 + q^{2} - 2q \right)\]
\[4 + 4q + 4q^{2} =\]
\[= 13 + 13q^{2} - 26q\ \ \]
\[9q^{2} - 30q + 9 = 0\ \ \ |\ :3\]
\[3q^{2} - 10q + 3 = 0\]
\[D = 100 - 36 = 64\]
\[q = \frac{10 + 8}{6} = 3;\ \ \]
\[q = \frac{10 - 8}{6} = \frac{1}{3}\]
\[b_{1}\left( 1 + q^{2} - 2q \right) = 8\ \ \]
\[b_{1}(1 - q)^{2} = 8\]
\[b_{1} = \frac{8}{(1 - q)^{2}}\]
\[при\ \ q = 3:\ \ \]
\[b_{1} = \frac{8}{(1 - 3)^{2}} = \frac{8}{4} = 2,\ \ \]
\[b_{2} = 2 \cdot 3 = 6,\ \ \]
\[b_{3} = 6 \cdot 3 = 18;\]
\[при\ q = \frac{1}{3}:\ \ \]
\[b_{1} = \frac{8}{\left( 1 - \frac{1}{3} \right)^{2}} = \frac{8 \cdot 9}{4} = 18,\]
\[\text{\ \ }b_{2} = 18 \cdot \frac{1}{3} = 6,\]
\[\text{\ \ }b_{3} = 6 \cdot \frac{1}{3} = 2.\]
\[Ответ:2,\ 6,\ 18\ \ \ или\ \ \ 18,\ 6,\ 2.\]
\[\boxed{\mathbf{864.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ 0,(21) = 0,2121\ldots = 0,21 +\]
\[+ 0,0021 + \ldots\]
\[q = 0,0021\ :0,21 = 0,01\]
\[S = \frac{b_{1}}{1 - q} = \frac{0,21}{1 - 0,01} = \frac{0,21}{0,99} =\]
\[= \frac{21}{99} = \frac{7}{33} \Longrightarrow \ \ \ \ 0,(21) = \frac{7}{33};\]
\[2)\ 0,2(3) = 0,2333\ldots = 0,2 +\]
\[+ 0,03 + 0,003 + \ldots = 0,2 + S\]
\[q = 0,003\ :0,03 = 0,1\]
\[S = \frac{0,03}{1 - 0,1} = \frac{0,03}{0,9} = \frac{3}{90} = \frac{1}{30};\ \]
\[\ \ \ \ 0,2(3) = \frac{1}{30} + 0,2 =\]
\[= \frac{1}{30} + \frac{2}{10} = \frac{1}{30} + \frac{6}{30} = \frac{7}{30};\]
\[3)\ 2,(7) = 2,777\ldots = 2 + 0,7 +\]
\[+ 0,07 + \ldots = 2 + S\]
\[q = 0,07\ :0,7 = 0,1\]
\[S = \frac{0,7}{1 - 0,1} = \frac{0,7}{0,9} = \frac{7}{9} \Longrightarrow\]
\[\Longrightarrow \ 2,(7) = 2 + \frac{7}{9} = 2\frac{7}{9};\]
\[4)\ 0,(19) = 0,191919\ldots =\]
\[= 0,19 + 0,0019 +\]
+\(0,000019 + \ldots\)
\[q = 0,0019\ :0,19 = 0,01\]
\[S = \frac{0,19}{1 - 0,01} = \frac{0,19}{0,99} = \frac{19}{99} \Longrightarrow \ \ 0,\]
\[(19) = \frac{19}{99};\]
\[5)\ 0,(901) = 0,901 +\]
\[+ 0,000901 + \ldots\]
\[q = 0,000901\ :0,901 = 0,001\]
\[S = \frac{0,901}{1 - 0,001} = \frac{0,901}{0,999} =\]
\[= \frac{901}{999} \Longrightarrow \ \ \ 0,(901) = \frac{901}{999};\]
\[6)\ 1,0(7) = 1,0777\ldots = 1 +\]
\[+ 0,07 + 0,007 + \ldots = 1 + S\]
\[q = 0,007\ :0,07 = 0,1\]
\[S = \frac{0,07}{1 - 0,1} = \frac{0,07}{0,9} = \frac{7}{90} \Longrightarrow\]
\[\Longrightarrow \ \ 1,0(7) = 1 + \frac{7}{90} = 1\frac{7}{90}.\]
