Решебник по алгебре 9 класс Мерзляк Задание 1041

\[\boxed{\mathbf{1041\ (1041).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ S = 4;\ \ q = \frac{1}{2}\]

\[S = \frac{b_{1}}{1 - q} \Longrightarrow b_{1} = S \cdot (1 - q) =\]

\[= 4 \cdot \left( 1 - \frac{1}{2} \right) = 4 \cdot \frac{1}{2} = 2;\]

\[2)\ S = \sqrt{2} + 1;\ \ q = \frac{\sqrt{2}}{2}\]

\[S = \frac{b_{1}}{1 - q} \Longrightarrow \ \]

\[b_{1} = S \cdot (1 - q) =\]

\[= \left( \sqrt{2} + 1 \right)\left( 1 - \frac{\sqrt{2}}{2} \right) =\]

\[= \left( \sqrt{2} + 1 \right) \cdot \frac{2 - \sqrt{2}}{2} =\]

\[= \frac{2\sqrt{2} - 2 + 2 - \sqrt{2}}{2} = \frac{\sqrt{2}}{2};\]

\[3)\ S = \frac{16}{3};\ \ S_{5} = 5,5\]

\[S = \frac{b_{1}}{1 - q} = \frac{16}{3},\]

\[\ \ 3b_{1} = 16 - 16q,\ \ \]

\[b_{1} = \frac{16}{3} \cdot (1 - q)\]

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

\[\frac{b_{1}(q^{5} - 1)}{q - 1} = \frac{11}{2}\]

\[\frac{\frac{16}{3} \cdot (1 - q)(q^{5} - 1)}{q - 1} = \frac{11}{2}\]

\[- \frac{16}{3} \cdot \left( q^{5} - 1 \right) = \frac{11}{2},\ \ \]

\[q^{5} - 1 = \frac{- 11 \cdot 3}{2 \cdot 16},\ \ \]

\[q^{5} - 1 = - \frac{33}{32}\]

\[q^{5} = 1 - \frac{33}{32},\ \ q^{5} = - \frac{1}{32},\]

\[\ \ q = - \frac{1}{2}\]

\[b_{1} = \frac{16}{3} \cdot \left( 1 + \frac{1}{2} \right) = \frac{16 \cdot 3}{3 \cdot 2} = 8.\]