ГДЗ по алгебре 9 класс Макарычев Задание 565

Авторы:
Год:2020-2021-2022
Тип:учебник
Нужно другое издание?

Задание 565

\[\boxed{\text{565\ (565).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x_{n} = 2n - 1\]

\[x_{1} = 2 - 1 = 1\]

\[x_{2} = 4 - 1 = 3\]

\[x_{3} = 6 - 1 = 5\]

\[x_{4} = 8 - 1 = 7\]

\[x_{5} = 10 - 1 = 9\]

\[x_{6} = 12 - 1 = 11;\]

\[\textbf{б)}\ x_{n} = n^{2} + 1\]

\[\ x_{1} = 1 + 1 = 2\]

\[x_{2} = 4 + 1 = 5\]

\[x_{3} = 9 + 1 = 10\]

\[x_{4} = 16 + 1 = 17\]

\[x_{5} = 25 + 1 = 26\]

\[x_{6} = 36 + 1 = 37\]

\[\textbf{в)}\ x_{n} = \frac{n}{n + 1}\]

\[x_{1} = \frac{1}{1 + 1} = \frac{1}{2}\]

\[x_{2} = \frac{2}{2 + 1} = \frac{2}{3}\]

\[x_{3} = \frac{3}{3 + 1} = \frac{3}{4}\]

\[x_{4} = \frac{4}{4 + 1} = \frac{4}{5}\]

\[x_{5} = \frac{5}{5 + 1} = \frac{5}{6}\]

\[x_{6} = \frac{6}{6 + 1} = \frac{6}{7}\ \]

\[\textbf{г)}\ x^{n} = ( - 1)^{n + 1} \cdot 2\ \]

\[x_{1} = ( - 1)^{2} \cdot 2 = 2\]

\[x_{2} = - 2\]

\[x_{3} = 2\]

\[x_{4} = - 2\]

\[x_{5} = 2\]

\[x_{6} = - 2;\]

\[\textbf{д)}\ x_{n} = 2^{n - 3}\ \]

\[x_{1} = 2^{- 2} = \frac{1}{4}\]

\[x_{2} = 2^{- 1} = \frac{1}{2}\]

\[x_{3} = 2^{0} = 1\]

\[x_{4} = 2^{1} = 2\]

\[x_{5} = 2^{2} = 4\]

\[x_{6} = 2³ = 8;\]

\[\textbf{е)}\ x^{n} = 0,5 \cdot 4^{n}\ \]

\[x_{1} = 0,5 \cdot 4 = 2\]

\[x_{2} = 0,5 \cdot 4^{2} = 8\]

\[x_{3} = 0,5 \cdot 4^{3} = 32\]

\[x_{4} = 0,5 \cdot 4^{4} = 128\]

\[x_{5} = 0,5 \cdot 4^{5} = 512\]

\[x_{6} = 0,5 \cdot 4^{6} = 2048.\]

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам