Решебник по алгебре 9 класс Макарычев Задание 998

\[\boxed{\text{998\ (998).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[q > 0,\ \ b_{1} \cdot b_{2} = \frac{1}{27},\ \ \]

\[b_{3} \cdot b_{4} = 3,\ \ S_{4} = ?\]

\[b_{1}^{2}q = \frac{1}{27},\ \ b_{1}^{2}q^{5} = 3,\ \ \]

\[q^{4} = 3\ :\frac{1}{27} = 81,\ \ q = 3,\]

\[b_{1} = \sqrt{\frac{1}{27}\ :3} = \sqrt{\frac{1}{81}} = \pm \frac{1}{9}.\]

\[S_{4} = b_{1} \cdot \frac{q^{4} - 1}{q - 1},\]

\[1)\ S_{4} = \frac{1}{9} \cdot \ \frac{3^{4} - 1}{3 - 1} = \frac{40}{9} = 4\frac{4}{9}.\]

\[2)\ S_{4} = - \frac{1}{9} \cdot \frac{3^{4} - 1}{3 - 1} = - \frac{40}{9} =\]

\[= - 4\frac{4}{9}.\]

\[Ответ:4\frac{4}{9}\ или\ \ \left( - 4\frac{4}{9} \right).\]