ГДЗ по алгебре 9 класс Макарычев Задание 701

\[\boxed{\text{701}\text{\ (701)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{3} = 225,\ \ b_{4} = - 135,\]

\[b_{5} = 81,\]

\[q = \frac{b_{4}}{b_{3}} = - \frac{135}{225} = - \frac{3}{5} = - 0,6\]

\[b_{3} = b_{1} \cdot q^{2}\]

\[b_{1} = \frac{b_{3}}{q^{2}} = 225\ :\left( - \frac{3}{5} \right)^{2} =\]

\[= 225 \cdot \frac{25}{9} = 625\]

\[b_{2} = b_{1} \cdot q = 625 \cdot \left( - \frac{3}{5} \right) =\]

\[= - 375\]

\[b_{6} = b_{5} \cdot q = 81 \cdot \left( - \frac{3}{5} \right) =\]

\[= - \frac{243}{5} = - 48,6.\]

\[\textbf{б)}\ b_{4} = 36,\ \ b_{5} = 54,\]

\[q = \frac{b_{5}}{b_{4}} = \frac{54}{36} = 1,5\]

\[b_{3} = \frac{b_{4}}{q} = 36\ :\frac{3}{2} = 24\]

\[b_{2} = \frac{b_{3}}{q} = 24\ :\frac{3}{2} = 16\]

\[b_{1} = \frac{b_{2}}{q} = 16\ :\frac{3}{2} = \frac{32}{3} = 10\frac{2}{3}.\]