Страница 232 номер 856, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(b_{4} - b_{2} = 30\) и \(b_{4} - b_{3} = 24\)
\(\begin{cases} b_{4} - b_{2} = 30 \\ b_{4} - b_{3} = 24 \end{cases}\ \Rightarrow b_{3} - b_{2} = 6 \Rightarrow b_{1}q^{2} - b_{1}q = 6 \Rightarrow b_{1} = \dfrac{6}{q^{2} - q}\)
\(b_{4} - b_{2} = 30\)
\(b_{1}q^{3} - b_{1}q = 30\)
\(b_{1}\left( q^{3} - q \right) = 30\)
\(\dfrac{6}{q^{2} - q} \cdot q\left( q^{2} - 1 \right) = 30\)
\(6q(q - 1)(q + 1) = 30q(q - 1)\)
\(q + 1 = 5\)
\(q = 4\)
\(b_{1} = \dfrac{6}{4^{2} - 4} = \dfrac{6}{12} = 0{,}5\)

2) \(b_{2} - b_{5} = 78\) и \(b_{3} + b_{4} + b_{5} = - 117\)
\(b_{2} - b_{5} = 78\)
\(b_{1}q - b_{1}q^{4} = 78\)
\(b_{1}q\left( 1 - q^{3} \right) = 78\)
\(b_{1} = \dfrac{78}{q(1 - q^{3})}\)
\(b_{3} + b_{4} + b_{5} = - 117\)
\(b_{1}\left( q^{2} + q^{3} + q^{4} \right) = - 117\)
\(\dfrac{78}{q\left( 1 - q^{3} \right)} \cdot q^{2}\left( 1 + q + q^{2} \right) = - 117\)
\(\dfrac{78q(1 + q + q^{2})}{(1 - q)(1 + q + q^{2})} = - 117\)
\(78q = - 117(1 - q)\)
\(78q = - 117 + 117q\)
\(- 39q = - 117\)
\(q = 3\)
\(b_{1} = \dfrac{78}{3(1 - 3^{3})} = \dfrac{26}{1 - 27} = - 1\)

Ответ: 1) \(b_{1} = 0{,}5;\ q = 4\); 2) \(b_{1} = - 1;\ q = 3\).