Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по алгебре 9 класс Макарычев Задание 704
Задание 704
\[\boxed{\text{704\ (704).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x_{n} = 2^{n}\]
\[x_{1} = 2^{1} = 2\]
\[x_{2} = 2^{2} = 4\]
\[x_{3} = 2^{3} = 8\]
\[q = \frac{x_{3}}{x_{2}} = \frac{x_{2}}{x_{1}} =\]
\[= 2 \Longrightarrow геометрическая\ \]
\[прогрессия.\]
\[\textbf{б)}\ x_{n} = 3^{- n}\]
\[x_{1} = 3^{- 1} = \frac{1}{3}\]
\[x_{2} = 3^{- 2} = \frac{1}{9}\]
\[x_{3} = 3^{- 3} = \frac{1}{27}\]
\[q = \frac{x_{3}}{x_{2}} = \frac{x_{2}}{x_{1}} =\]
\[= \frac{1}{3} \Longrightarrow геометрическая\ \]
\[прогрессия.\]
\[\textbf{в)}\ x_{n} = n²\]
\[x_{1} = 1^{2} = 1\]
\[x_{2} = 2^{2} = 4\]
\[x_{3} = 3² = 9\]
\[q = \frac{x_{n}}{x_{n - 1}} = \frac{n^{2}}{(n - 1)^{2}} \Longrightarrow зависит\ \]
\[от\ n,\ не\ геометрическая\ \]
\[прогрессия.\]
\[\textbf{г)}\ x_{n} = ab^{n},\ \ где\ a \neq 0,\]
\[b \neq 0,\]
\[q = \frac{x_{n}}{x_{n - 1}} = \frac{ab^{n}}{ab^{n - 1}} = b^{n - n + 1} =\]
\[= b \Longrightarrow геометрическая\ \]
\[прогрессия.\]
\[\boxed{\text{704.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x² - x - 42 = (x - 7)(x + 6)\]
\[x^{2} - x - 42 = 0\]
\[D = 1 + 168 = 169\]
\[x_{1} = \frac{1 - 13}{2} = - 6,\ \ \]
\[x_{2} = \frac{1 + 13}{7} = 7.\]
\[\textbf{б)}\ y² + 9y + 18 =\]
\[= (y + 3)(y + 6)\]
\[y^{2} + 9y + 18 = 0\]
\[D = 81 - 72 = 9\]
\[y_{1} = \frac{- 9 + 3}{2} = - 3,\ \ \]
\[y_{2} = \frac{- 9 - 3}{2} = - 6.\]
\[\textbf{в)}\ 81x² + 18x + 1 =\]
\[= (9x + 1)^{2} = (9x + 1)(9x + 1)\]
\[\textbf{г)}\ 16b² - 24b + 9 =\]
\[= (4b - 3)^{2} = (4b - 3)(4b - 3)\]
\[\textbf{д)}\ 6x² - x - 1 =\]
\[= 6 \cdot \left( x - \frac{1}{2} \right)\left( x + \frac{1}{3} \right) =\]
\[= (2x - 1)(3x + 1)\]
\[6x² - x - 1 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{1 + 5}{12} = \frac{6}{12} = \frac{1}{2},\ \ \]
\[x_{2} = \frac{1 - 5}{12} = - \frac{4}{12} = - \frac{1}{3}.\]
\[\textbf{е)}\ 3a² - 13a - 10 =\]
\[= 3 \cdot (a - 5)\left( a + \frac{2}{3} \right) =\]
\[= (a - 5)(3a + 2)\]
\[3a^{2} - 13a - 10 = 0\]
\[D = 169 + 120 = 289\]
\[a_{1} = \frac{13 + 17}{6} = 5,\]
\[\text{\ \ }a_{2} = \frac{13 - 17}{6} = - \frac{4}{6} = - \frac{2}{3}.\]
