ГДЗ по алгебре 9 класс Мерзляк Задание 1037

\[\boxed{\mathbf{1037\ (1037).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ b_{5} = - \frac{16}{27},\ \ q = - \frac{2}{3},\]

\[\text{\ \ }b_{1} - ?\]

\[b_{5} = b_{1}q^{4}\text{\ \ }\]

\[b_{1} = \frac{b_{5}}{q^{4}} = \frac{- 16 \cdot 3^{4}}{27 \cdot 2^{4}} =\]

\[= \frac{- 16 \cdot 81}{27 \cdot 16} = - 3\]

\[Ответ:\ b_{1} = - 3.\]

\[2)\ b_{1} = \frac{2}{3},\ \ b_{4} = \frac{9}{32},\ \ q - ?\]

\[b_{4} = b_{1}q^{3},\ \ q^{3} = \frac{b_{4}}{b_{1}},\]

\[\text{\ \ }q^{3} = \frac{9 \cdot 3}{32 \cdot 2},\ \ q^{3} = \frac{27}{64},\]

\[\ \ q = \frac{3}{4}\]

\[\mathbf{Ответ}:q = \frac{3}{4}\mathbf{.}\]

\[3)\ b_{7} = 192,\ \ q = 2,\ \ S_{7} - ?\]

\[b_{7} = b_{1}q^{6},\ \ \]

\[b_{1} = \frac{b_{7}}{q^{6}} = \frac{192}{64} = 3\]

\[S_{7} = \frac{b_{1}\left( q^{7} - 1 \right)}{q - 1} = \frac{3 \cdot \left( 2^{7} - 1 \right)}{2 - 1} =\]

\[= 3 \cdot (128 - 1) = 3 \cdot 127 = 381\]

\[Ответ:381.\]

\[4)\ b_{5} = 9\sqrt{6},\ \ q = \sqrt{3},\ \ \]

\[S_{5} - ?\]

\[b_{5} = b_{1}q^{4},\ \ \]

\[b_{1} = \frac{b_{5}}{q^{4}} = \frac{9\sqrt{6}}{9} = \sqrt{6}\]

\[S_{5} = \frac{b_{1}\left( q^{5} - 1 \right)}{q - 1} =\]

\[= \frac{\sqrt{6} \cdot \left( 9\sqrt{3} - 1 \right)}{\sqrt{3} - 1} =\]

\[= \frac{\sqrt{6} \cdot \left( 9\sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)} =\]

\[= \frac{\sqrt{6} \cdot \left( 27 + 9\sqrt{3} - \sqrt{3} - 1 \right)}{3 - 1} =\]

\[= \frac{\sqrt{6} \cdot \left( 26 + 8\sqrt{3} \right)}{2} =\]

\[= \sqrt{6} \cdot \left( 13 + 4\sqrt{3} \right) =\]

\[= 13\sqrt{6} + 4\sqrt{18} = 13\sqrt{6} + 12\sqrt{2}\]

\[Ответ:13\sqrt{6} + 12\sqrt{2}.\]