Решебник по алгебре 9 класс Макарычев Задание 706

 

\[\boxed{\text{706\ (706).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\text{706.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{3x + 2y}{x}\ \]

\[\frac{2x + 3y}{y} = 7\]

\[\frac{2x}{y} + 3 = 7\]

\[\frac{2x}{y} = 4\]

\[\frac{x}{y} = 2;\]

\[\frac{3x + 2y}{x} = 3 + \frac{2y}{x} = 3 + 2 \cdot \frac{1}{2} =\]

\[= 4.\]

\[\textbf{б)}\ \frac{b}{a + b}\ \]

\[\frac{4a - 5b}{b} = 3\]

\[4a - 5b = 3b\]

\[4a = 8b\]

\[a = 2b,\]

\[\frac{b}{a + b} = \frac{b}{2b + b} = \frac{1}{3}.\]