Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1258

Авторы:
Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 1258

\[\boxed{\mathbf{1258}\mathbf{.}}\]

\[1)\ 0,(4):\]

\[x = 0,(4)\]

\[10x = 4,(4)\]

\[x = \frac{9x}{9} = \frac{10x - x}{9} =\]

\[= \frac{4,(4) - 0,(4)}{9} = \frac{4}{9}.\]

\[Ответ:\ \ \frac{4}{9}.\]

\[2)\ 2,(7):\]

\[x = 2,(7)\]

\[10x = 27,(7)\]

\[x = \frac{9x}{x} = \frac{10x - x}{9} =\]

\[= \frac{27,(7) - 2,(7)}{9} = \frac{25}{9} = 2\frac{7}{9}.\]

\[Ответ:\ \ 2\frac{7}{9}.\]

\[3)\ 0,(21):\]

\[x = 0,(21)\]

\[100x = 21,(21)\]

\[x = \frac{99x}{99} = \frac{100x - x}{99} =\]

\[= \frac{21,(21) - 0,(21)}{99} = \frac{21}{99} = \frac{7}{33}.\]

\[Ответ:\ \ \frac{7}{33}.\]

\[4)\ 1,(36):\]

\[x = 1,(36)\]

\[100x = 136,(36)\]

\[x = \frac{99x}{99} = \frac{100x - x}{99} =\]

\[= \frac{136,(36) - 1,(36)}{99} =\]

\[= \frac{135}{99} = \frac{15}{11} = 1\frac{4}{11}.\]

\[Ответ:\ \ 1\frac{4}{11}.\]

\[5)\ 0,3(5):\]

\[x = 0,3(5)\]

\[10x = 3,(5)\]

\[100x = 35,(5)\]

\[x = \frac{90x}{90} = \frac{100x - 10x}{90} =\]

\[= \frac{35,(5) - 3,(5)}{90} = \frac{32}{90}.\]

\[Ответ:\ \ \frac{32}{90}.\]

\[6)\ 0,21(3):\]

\[x = 0,21(3)\]

\[100x = 21,(3)\]

\[1000x = 213,(3)\]

\[x = \frac{900x}{900} = \frac{1000x - 100x}{900} =\]

\[= \frac{213,(3) - 21,(3)}{900} = \frac{192}{900} = \frac{16}{75}.\]

\[Ответ:\ \ \frac{16}{75}.\]

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам