Страница 6 номер 3, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Записать в виде обыкновенной дроби бесконечную десятичную дробь: 1) \(0{,}(6)\); 2) \(1{,}(55)\); 3) \(0{,}1(2)\); 4) \(-0{,}(8)\); 5) \(-3{,}(27)\); 6) \(-2{,}3(82)\).
1) \(0{,}(6)\): Пусть \(x = 0{,}(6)\), тогда \(10x = 6{,}(6)\). \(10x - x = 6{,}(6) - 0{,}(6)\), \(9x = 6\), \(x = \frac{6}{9} = \frac{2}{3}\);
2) \(1{,}(55)\): Пусть \(x = 1{,}(55)\). Тогда \(100x = 155{,}(55)\).
\(100x - x = 155{,}(55) - 1{,}(55), 99x = 154,\)
\(x = \dfrac{154}{99} = \dfrac{14 \cdot 11}{9 \cdot 11} = \dfrac{14}{9} = 1\dfrac{5}{9};\)
3) \(0{,}1(2)\): Пусть \(x = 0{,}1(2)\), тогда \(10x = 1{,}(2)\) и \(100x = 12{,}(2)\).
\(100x - 10x = 12{,}(2) - 1{,}(2), 90x = 11, x = \dfrac{11}{90};\)
4) \(-0{,}(8)\): Пусть \(x = -0{,}(8)\), тогда \(10x = -8{,}(8)\).
\(10x - x = -8{,}(8) - \left(-0{,}(8)\right), 9x = -8, x = -\dfrac{8}{9}\)
5) \(-3{,}(27)\): Пусть \(x = -3{,}(27)\), тогда \(100x = -327{,}(27)\).
\(100x - x = -327{,}(27) - \left(-3{,}(27)\right),\)
\(99x = -324, x = -\dfrac{324}{99} = -\dfrac{9 \cdot 36}{9 \cdot 11} =\)
\(= -\dfrac{36}{11} = -3\dfrac{3}{11}\)
6) \(-2{,}3(82)\): Пусть \(x = -2{,}3(82)\), тогда
\(10x = -23{,}(82)\) и \(1000x = -2382{,}(82)\).
\(1000x - 10x = -2382{,}(82) - \left(-23{,}(82)\right)\)
\(990x = -2359, x = -\dfrac{2359}{990} = -2\dfrac{379}{990}\)
