Решебник по алгебре 9 класс Мерзляк Задание 855

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\[1)\left\{ \begin{matrix} b_{5} = 3b_{3}\text{\ \ \ \ \ \ \ \ } \\ b_{6} - b_{2} = 48 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} b_{1} \cdot q^{4} = 3b_{1}q^{2} \\ b_{1}q^{5} - b_{1}q = 48 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} b_{1}q^{4} - 3b_{1}q^{2} = 0 \\ b_{1}q^{5} - b_{1}q = 48 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} b_{1}q^{2}\left( q^{2} - 3 \right) = 0 \\ b_{1}q\left( q^{4} - 1 \right) = 48 \\ \end{matrix} \right.\ \]

\[b_{1}q \neq 0;\ \ тогда\ \ q^{2} =\]

\[= 3 \Longrightarrow \ \ q = \pm \sqrt{3}\]

\[b_{1}q\left( \left( \pm \sqrt{3} \right)^{4} - 1 \right) = 48\ \ \]

\[b_{1}q(9 - 1) = 48\ \]

\[b_{1}q = 6 \Longrightarrow b_{1} = \frac{6}{q}\]

\[при\ \ q = \sqrt{3}:\ \ \ \ b_{1} = \frac{6}{\sqrt{3}} =\]

\[= \frac{6\sqrt{3}}{3} = 2\sqrt{3};\ \]

\[при\ \ q = - \sqrt{3}:\ \ b_{1} = \frac{6}{- \sqrt{3}} =\]

\[= \frac{6\sqrt{3}}{- 3} = - 2\sqrt{3}.\]

\[Ответ:\ b_{1} = 2\sqrt{3};\ q = \sqrt{3}\text{\ \ }или\ \ \ \]

\[b_{1} = - 2\sqrt{3};\ \ q = - \sqrt{3}.\]

\[2)\ b_{4} + b_{7} = \frac{56}{9}\text{\ \ \ \ }и\ \ \ \]

\[b_{5} - b_{6} + b_{7} = \frac{14}{9}\]

\[\left\{ \begin{matrix} b_{1}q^{3} + b_{1}q^{6} = \frac{56}{9}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ b_{1}q^{4} - b_{1}q^{5} + b_{1}q^{6} = \frac{14}{9} \\ \end{matrix} \right.\ \ \]

\[\text{\ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q^{3}\left( 1 + q^{3} \right) = \frac{56}{9}\text{\ \ \ \ \ \ \ \ \ } \\ b_{1}q^{4}\left( 1 - q + q^{2} \right) = \frac{14}{9} \\ \end{matrix} \right.\ \ \ |(\ :)\]

\[\frac{b_{1}q^{3}\left( 1 + q^{3} \right)}{b_{1}q^{4}\left( 1 - q + q^{2} \right)} = \frac{56 \cdot 9}{9 \cdot 14}\ \]

\[\frac{(1 + q)(1 - q + q^{2})}{q(1 - q + q^{2})} = 4\]

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

\[4q = 1 + q\ \ \]

\[3q = 1 \Longrightarrow \ \ q = \frac{1}{3}\]

\[b_{1}\left( q^{3} + q^{6} \right) = \frac{56}{9}\text{\ \ }\]

\[b_{1} = \frac{56}{9} \cdot \left( \left( \frac{1}{3} \right)^{3} + \left( \frac{1}{3} \right)^{6} \right) =\]

\[= \frac{56 \cdot 27 \cdot 27}{9 \cdot 28} = 162\]

\[Ответ:\ b_{1} = 162;\ q = \frac{1}{3}.\]

\[3)\ b_{5} - b_{4} = 168;\ \ b_{3} + b_{4} = - 28\]

\[\ \left\{ \begin{matrix} b_{1}q^{4} - b_{1}q^{3} = 168 \\ b_{1}q^{2} + b_{1}q^{3} = - 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1}q^{3}(q - 1) = 168 \\ b_{1}q^{2}(1 + q) = - 28\ \\ \end{matrix} \right.\ \ \ \ |(\ :)\]

\[\frac{b_{1}q^{3}(q - 1)}{b_{1}q^{2}(1 + q)} = \frac{168}{- 28}\]

\[\frac{q(q - 1)}{q + 1} = - 6\]

\[- 6q - 6 = q(q - 1)\text{\ \ }\]

\[- 6q - 6 = q^{2} - q\ \ \]

\[q^{2} + 5q + 6 = 0\]

\[q_{1} + q_{2} = - 5;\ \ \ q_{1} = - 2\]

\[q_{1}q_{2} = 6;\ \ \ \ \ \ \ \ \ \ \ \ q_{2} = - 3\]

\[при\ \ q = - 2:\ \ \ \]

\[b_{1} \cdot ( - 2)^{2} + b_{1} \cdot ( - 2)^{3} = - 28\ \ \]

\[4b_{1} + ( - 8)b_{1} = - 28\]

\[- 4b_{1} = - 28 \Longrightarrow \ \ b_{1} = 7;\]

\[при\ q = - 3:\ \ \]

\[b_{1} \cdot ( - 3)^{2} + b_{1} \cdot ( - 3)^{3} = - 28\ \ \]

\[9b_{1} - 27b_{1} = - 28\]

\[- 18b_{1} = - 28 \Longrightarrow \ \ b_{1} = \frac{14}{9}.\]

\[\mathbf{Ответ:\ }b_{1} = 7;\ \ q = - 2\ \ \ или\ \]

\[\text{\ \ \ }b_{1} = \frac{14}{9};\ \ q = - 3.\]