Решебник по алгебре 9 класс Макарычев Задание 673

 

\[\boxed{\text{673}\text{\ (673)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{3} = - 19,\ \ a_{4} = - 11,5;\ \ \]

\[d = - 11,5 + 19 = 7,5;\]

\[a_{2} = a_{3} - 7,5 = - 19 - 7,5 = - 26,5;\]

\[a_{1} = a_{2} - 7,5 = - 26,5 - 7,5 = - 34;\]

\[a_{5} = a_{4} + d = - 11,5 + 7,5 = - 4;\]

\[\textbf{б)}\ a_{2} = - 8,5,\ \ a_{4} = - 4,5,\ \ \]

\[2d = a_{4} - a_{2} = - 4,5 + 8,5 = 4 \Longrightarrow d = 2;\]

\[a_{1} = a_{2} - d = - 8,5 - 2 = - 10,5;\]

\[a_{3} = a_{2} + d = - 8,5 + 2 = - 6,5;\]

\[a_{5} = a_{4} + d = - 4,5 + 2 = - 2,5;\]

\[a_{6} = a_{5} + d = - 2,5 + 2 = - 0,5.\]

\[\boxed{\text{673.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[c_{1} = 135,\ \ c_{9} = \frac{5}{3},\]

\[c_{9} = c_{1} \cdot q^{8},\]

\[q^{8} = \frac{5}{3 \cdot 135} = \frac{1}{81},\]

\[q = \pm \frac{1}{\sqrt{3}},\]

\[1)\ c_{2} = c_{1} \cdot q = \frac{135}{\sqrt{3}} = 45\sqrt{3},\]

\[c_{3} = c_{2} \cdot q = \frac{45\sqrt{3}}{\sqrt{3}} = 45,\]

\[c_{4} = c_{3} \cdot q = \frac{45}{\sqrt{3}} = 15\sqrt{3},\]

\[c_{5} = c_{4} \cdot q = \frac{15\sqrt{3}}{\sqrt{3}} = 15,\]

\[c_{6} = c_{5} \cdot q = \frac{15}{\sqrt{3}} = 5\sqrt{3}\]

\[c_{7} = c_{6} \cdot q = \frac{5\sqrt{3}}{\sqrt{3}} = 5,\]

\[c_{8} = c_{7} \cdot q = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}.\]

\[2)\ c_{2} = c_{1} \cdot q = \frac{135}{- \sqrt{3}} = - 45\sqrt{3},\]

\[c_{3} = c_{2} \cdot q = \frac{- 45\sqrt{3}}{- \sqrt{3}} = 45,\]

\[c_{4} = c_{3} \cdot q = \frac{45}{- \sqrt{3}} = - 15\sqrt{3},\]

\[c_{5} = c_{4} \cdot q = \frac{- 15\sqrt{3}}{- \sqrt{3}} = 15,\]

\[c_{6} = c_{5} \cdot q = \frac{15}{- \sqrt{3}} = - 5\sqrt{3},\]

\[c_{7} = c_{6} \cdot q = \frac{- 5\sqrt{3}}{- \sqrt{3}} = 5,\]

\[c_{8} = c_{7} \cdot q = \frac{5}{- \sqrt{3}} = - \frac{5\sqrt{3}}{3}.\]