Страница 232 номер 855, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(b_{5} = 3b_{3}\) и \(b_{6} - b_{2} = 48\)
\(\begin{cases} b_{5} = 3b_{3} \\ b_{6} - b_{2} = 48 \end{cases}\ \Rightarrow \begin{cases} b_{1}q^{4} = 3b_{1}q^{2} \\ b_{1}q^{5} - b_{1}q = 48 \end{cases}\ \Rightarrow \begin{cases} q^{2} = 3 \\ b_{1} = \frac{48}{q(q^{4} - 1)} \end{cases}\)
\(\begin{cases} q = \sqrt{3} \\ b_{1} = \frac{48}{\sqrt{3}\left( \left( \sqrt{3} \right)^{4} - 1 \right)} \end{cases}\ \Rightarrow \begin{cases} q = \sqrt{3} \\ b_{1} = \frac{48}{\sqrt{3} \cdot 8} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \end{cases}\)
\(\begin{cases} q = - \sqrt{3} \\ b_{1} = \frac{48}{- \sqrt{3}\left( \left( - \sqrt{3} \right)^{4} - 1 \right)} \end{cases}\ \Rightarrow \begin{cases} q = - \sqrt{3} \\ b_{1} = \frac{48}{- \sqrt{3} \cdot 8} = \frac{6}{- \sqrt{3}} = - 2\sqrt{3} \end{cases}\)

2) \(b_{4} + b_{7} = \frac{56}{9}\) и \(b_{5} - b_{6} + b_{7} = \frac{14}{9}\)
\(\begin{cases} b_{4} + b_{7} = \frac{56}{9} \\ b_{5} - b_{6} + b_{7} = \frac{14}{9} \end{cases}\ \Rightarrow \begin{cases} b_{1}q^{3} + b_{1}q^{6} = \frac{56}{9} \\ b_{1}q^{4} - b_{1}q^{5} + b_{1}q^{6} = \frac{14}{9} \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} b_{1}q^{3}\left( 1 + q^{3} \right) = \frac{56}{9} \\ b_{1}q^{4}\left( 1 - q + q^{2} \right) = \frac{14}{9} \end{cases}\ \Rightarrow \begin{cases} b_{1}q^{3}(1 + q)\left( 1 - q + q^{2} \right) = \frac{56}{9} \\ b_{1}q^{4}(1 - q + q^{2}) = \frac{14}{9} \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} \frac{14}{9} \cdot \frac{1}{q}(1 + q) = \frac{56}{9} \\ b_{1}q^{3}\left( 1 - q + q^{2} \right) = \frac{14}{9} \cdot \frac{1}{q} \end{cases}\ \Rightarrow \begin{cases} 1 + q = 4q \\ b_{1}q^{3}\left( 1 - q + q^{2} \right) = \frac{14}{9} \cdot \frac{1}{q} \end{cases}\ \Rightarrow q = \dfrac{1}{3}\)
\(b_{1}q^{3}\left( 1 + q^{3} \right) = \dfrac{56}{9} \Rightarrow b_{1} = \dfrac{56}{9} \cdot \dfrac{1}{q^{3}\left( 1 + q^{3} \right)} = \dfrac{56}{9} \cdot \dfrac{1}{\dfrac{1}{27}\left( 1 + \dfrac{1}{27} \right)} =\)
\(= \dfrac{56}{9} \cdot \dfrac{27 \cdot 27}{28} = \dfrac{2 \cdot 3 \cdot 27}{1} = 162\)

3) \(b_{5} - b_{4} = 168\) и \(b_{3} + b_{4} = - 28\)
\(\begin{cases} b_{5} - b_{4} = 168 \\ b_{3} + b_{4} = - 28 \end{cases}\ \Rightarrow b_{5} + b_{3} = 140 \Rightarrow\)
\(\Rightarrow b_{4}q + \dfrac{b_{4}}{q} = 140 \Rightarrow b_{4}\left( \dfrac{q^{2} + 1}{q} \right) = 140 \Rightarrow b_{4} = \dfrac{140q}{q^{2} + 1} = b_{1}q^{3}\)
\(b_{1} = \dfrac{140q}{q^{3}(q^{2} + 1)}\)
\(b_{4}q - b_{4} = 168\)
\(b_{4}(q - 1) = 168\)
\(\dfrac{140q}{q^{2} + 1}(q - 1) = 168\)
\(140q(q - 1) = 168\left( q^{2} + 1 \right)\)
\(140q^{2} - 140q = 168q^{2} + 168\)
\(28q^{2} + 140q + 168 = 0\)
\(q^{2} + 5q + 6 = 0\)
\((q + 2)(q + 3) = 0\)
\(q = - 2 \Rightarrow b_{1} = \dfrac{140q}{q^{3}(q^{2} + 1)} = \dfrac{140 \cdot ( - 2)}{- 8 \cdot 5} = 7\)
\(q = - 3 \Rightarrow b_{1} = \dfrac{140q}{q^{3}(q^{2} + 1)} = \dfrac{140 \cdot ( - 3)}{- 27 \cdot 10} = \dfrac{14}{9}\)

Ответ: 1) \(q = \sqrt{3};\ b_{1} = 2\sqrt{3}\); \(q = - \sqrt{3};\ b_{1} = - 2\sqrt{3}\); 2) \(q = \frac{1}{3};\ b_{1} = 162\); 3) \(q = - 2 \Rightarrow b_{1} = 7\); \(q = - 3 \Rightarrow b_{1} = \frac{14}{9}\).