Страница 219 номер 702, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(S_{3} = \frac{b_{1}(1 - q^{3})}{1 - q} = 372\),
\(\dfrac{12(1 - q)\left(1 + q + q^{2}\right)}{1 - q} = 372,\)
\(1 + q + q^{2} = 31,\)
\(q^{2} + q - 30 = 0,\)
\(q_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 120}}{2},\)
\(q_{1} = 5, q_{2} = -6;\)

2) \(S_{3} = \frac{b_{1}(1 - q^{3})}{1 - q} = 157\),
\(1 + q + q^{2} = 157,\)
\(q^{2} + q - 156 = 0,\)
\(q_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 624}}{2},\)
\(q_{1} = 12, q_{2} = -13;\)

3) \(S_{3} = \frac{b_{1}(1 - q^{3})}{1 - q} = \frac{b_{1}q^{2}(1 - q^{3})}{q^{2}(1 - q)} = 372\),
\(\dfrac{300\left(1 + q + q^{2}\right)}{q^{2}} = 372,\)
\(25\left(1 + q + q^{2}\right) = 31q^{2},\)
\(25 + 25q + 25q^{2} - 31q^{2} = 0,\)
\(6q^{2} - 25q - 25 = 0,\)
\(q_{1{,}2} = \dfrac{25 \pm \sqrt{625 + 600}}{12},\)
\(q_{1} = 5, q_{2} = -\dfrac{5}{6};\)

4) \(S_{3} = \frac{b_{1}(1 - q^{3})}{1 - q} = 157\),
\(144\left(1 + q + q^{2}\right) = 157,\)
\(q^{2} + q - \dfrac{13}{144} = 0,\)
\(q_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + \frac{52}{144}}}{2} = \dfrac{-1 \pm \frac{7}{6}}{2},\)
\(q_{1} = \dfrac{1}{12}, q_{2} = -\dfrac{13}{12} = -1\dfrac{1}{12}.\)