Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 189

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\[b_{1} + b_{4} = 27;\ \ b_{2} + b_{3} = 18:\]

\[\left\{ \begin{matrix} b_{1} + b_{1}q^{3} = 27\ \ \\ b_{1}q + b_{1}q^{2} = 18 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b_{1}\left( 1 + q^{3} \right) = 27 \\ b_{1}\left( q + q^{2} \right) = 18 \\ \end{matrix} \right.\ \text{\ \ \ }(\ :)\]

\[\frac{1 + q^{3}}{q + q^{2}} = \frac{27}{18}\]

\[\frac{(1 + q)\left( 1 - q + q^{2} \right)}{q(1 + q)} = \frac{3}{2}\]

\[\frac{1 - q + q^{2}}{q} = \frac{3}{2}\]

\[2 - 2q + 2q^{2} = 3q\]

\[2q^{2} - 5q + 2 = 0\]

\[D = 25 - 16 = 9\]

\[q_{1} = \frac{5 + 3}{4} = 2;\ \ q_{2} = \frac{5 - 3}{4} = \frac{1}{2}.\]

\[1)\ q_{1} = 2:\]

\[b_{1} \cdot \left( 1 + 2^{3} \right) = 27\]

\[9b_{1} = 27\]

\[b_{1} = 3.\]

\[2)\ q_{2} = \frac{1}{2}:\]

\[b_{1} \cdot \left( 1 + \left( \frac{1}{2} \right)^{3} \right) = 27\]

\[b_{1} \cdot \left( 1 + \frac{1}{8} \right) = 27\]

\[b_{1} \cdot \frac{9}{8} = 27\]

\[b_{1} = 27 \cdot \frac{8}{9}\]

\[b_{1} = 24.\]

\[Ответ:q = 2;b_{1} = 3\ \ или\]

\[\ q = \frac{1}{2};b_{1} = 24.\]