ГДЗ по алгебре 9 класс Мерзляк Задание 856

\[\boxed{\mathbf{856\ (856).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\left\{ \begin{matrix} b_{4} - b_{2} = 30 \\ b_{4} - b_{3} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1}q^{3} - b_{1}q = 30 \\ b_{1}q^{3} - b_{1}q^{2} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1}q\left( q^{2} - 1 \right) = 30 \\ b_{1}q^{2}(q - 1) = 24 \\ \end{matrix} \right.\ \ \ \ \ |\ :\]

\[\frac{b_{1}q\left( q^{2} - 1 \right)}{b_{1}q^{2}(q - 1)} = \frac{30}{24}\text{\ \ }\]

\[\frac{(q - 1)(q + 1)}{q(q - 1)} = \frac{5}{4}\text{\ \ }\]

\[\frac{q + 1}{q} = \frac{5}{4}\]

\[5q = 4q + 4 \Longrightarrow \ \ q = 4.\]

\[b_{1} \cdot 4 \cdot \left( 4^{2} - 1 \right) = 30\ \ \]

\[b_{1} \cdot 4 \cdot 15 = 30\ \ \]

\[b_{1} = \frac{30}{60} \Longrightarrow \ \ b_{1} = 0,5.\]

\[Ответ:\ b_{1} = 0,5;\ \ \ q = 4.\]

\[2)\ \left\{ \begin{matrix} b_{2} - b_{5} = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{3} + b_{4} + b_{5} = - 117 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b_{1}q - b_{1}q^{4} = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2} + b_{1}q^{3} + b_{1}q^{4} = - 117 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} b_{1}q\left( 1 - q^{3} \right) = 78\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( 1 + q + q^{2} \right) = - 117 \\ \end{matrix} \right.\ \ \ \ |\ :\]

\[\frac{b_{1}q\left( 1 - q^{3} \right)}{b_{1}q^{2}\left( 1 + q + q^{2} \right)} = \frac{78}{- 117}\]

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

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

\[3 \cdot (1 - q) = - 2q\ \ \]

\[3 - 3q = - 2q \Longrightarrow \ \ q = 3.\]

\[b_{1} \cdot 3 \cdot \left( 1 - 3^{3} \right) = 78\ \ \]

\[b_{1} \cdot 3 \cdot ( - 26) = 78\ \]

\[b_{1} = - \frac{78}{78} = - 1.\]

\[Ответ:\ b_{1} = - 1;\ q = 3.\]