Решебник по алгебре 9 класс Макарычев Задание 702

 

\[\boxed{\text{702\ (702).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x_{1} + 1;x_{2} + 1;\ldots;x_{n} + 1;\ldots;\]

\[c_{1} = x_{1} + 1,\ \ \]

\[c_{n} = x_{n} + 1 = x_{n} \cdot q^{n} + 1,\ \]

\[q = \frac{c_{n}}{c_{n - 1}} = \frac{x_{n} \cdot q^{n} + 1}{x_{n - 1} \cdot q^{n - 1} + 1} \Longrightarrow\]

\[\Longrightarrow зависит\ от\ n \Longrightarrow не\ \]

\[геометрическая\ прогрессия.\]

\[\textbf{б)}\ 3x_{1};3x_{2};\ldots;3x_{n};\ldots;\]

\[c_{1} = 3x_{1},\ \ c_{n} = 3x_{n},\ \ \]

\[\frac{c_{n}}{c_{n - 1}} = \frac{3x_{n}}{3x_{n - 1}} =\]

\[= q \Longrightarrow геометрическая\ \]

\[прогрессия.\]

\[\textbf{в)}\ x_{1}^{2};x_{2}^{2};\ldots;x_{n}^{2};\ldots;\]

\[c_{1} = x_{1}^{2},\ \ c_{n} = x_{n}^{2},\ \ \]

\[\frac{c_{n}}{c_{n - 1}} = \frac{x_{n}^{2}}{x_{(n - 1)}^{2}} =\]

\[= q^{2} \Longrightarrow геометрическая\ \]

\[прогресссия.\]

\[\textbf{г)}\frac{1}{x_{1}};\frac{1}{x_{2}};\ldots;\frac{1}{x_{n}};\ldots;\ \]

\[c_{1} = \frac{1}{x_{1}},\ \ c_{n} = \frac{1}{x_{n}} = \frac{1}{c_{1}q^{n - 1}},\ \ \]

\[\frac{c_{n}}{c_{n - 1}} = \frac{1}{c_{1}q^{n - 1}} \cdot c_{1}q^{n - 2} =\]

\[= q^{- 1} \Longrightarrow геометрическая\ \]

\[прогресссия.\]

\[\boxed{\text{702.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 12x³ - 3x^{2}y - 18xy^{2} =\]

\[= 3x(4x^{2} - xy - 6y^{2})\]

\[\textbf{б)}\ 42a^{5} - 6a^{4} + 30a^{3} =\]

\[= 6a^{3} \cdot (7a^{2} - a + 5)\]

\[\textbf{в)}\ 8ab - 14a - 12b + 21 =\]

\[= 4b(2a - 3) - 7 \cdot (2a - 3) =\]

\[= (2a - 3)(4b - 7)\]

\[\textbf{г)}\ x² - 5x - 9xy + 45y =\]

\[= x(x - 9y) - 5 \cdot (x - 9y) =\]

\[= (x - 9y)(x - 5).\]