Упр.15.15 ГДЗ Рабочая тетрадь Зубарева Мордкович 6 класс Экзамен (Математика)
- Выполните деление, рассуждая, как в предыдущем примере:
а) $$16\frac{8}{9}:8$$; $$34\frac{4}{9}:2$$; $$35\frac{5}{8}:5$$;
б) $$120\frac{6}{7}:3$$; $$14\frac{7}{3}:7$$; $$26\frac{8}{11}:2$$;
в) $$64\frac{16}{21}:4$$; $$38\frac{6}{11}:2$$; $$100\frac{50}{109}:25$$;
г) $$49\frac{14}{15}:7$$; $$84\frac{42}{271}:42$$; $$100\frac{75}{647}:25$$.
Выполним деление целой и дробной частей отдельно:
а)
$$16\frac{8}{9}:8=16:8+\frac{8}{9}:8=2+\frac{8}{9}\cdot\frac{1}{8}=2+\frac{1}{9}=2\frac{1}{9}$$
$$34\frac{4}{9}:2=34:2+\frac{4}{9}:2=17+\frac{4}{9}\cdot\frac{1}{2}=17+\frac{2}{9}=17\frac{2}{9}$$
$$35\frac{5}{8}:5=35:5+\frac{5}{8}:5=7+\frac{5}{8}\cdot\frac{1}{5}=7+\frac{1}{8}=7\frac{1}{8}$$
б)
$$120\frac{6}{7}:3=120:3+\frac{6}{7}:3=40+\frac{6}{7}\cdot\frac{1}{3}=40+\frac{2}{7}=40\frac{2}{7}$$
$$14\frac{7}{3}:7=14:7+\frac{7}{3}:7=2+\frac{7}{3}\cdot\frac{1}{7}=2+\frac{1}{3}=2\frac{1}{3}$$
$$26\frac{8}{11}:2=26:2+\frac{8}{11}:2=13+\frac{8}{11}\cdot\frac{1}{2}=13+\frac{4}{11}=13\frac{4}{11}$$
в)
$$64\frac{16}{21}:4=64:4+\frac{16}{21}:4=16+\frac{16}{21}\cdot\frac{1}{4}=16+\frac{4}{21}=16\frac{4}{21}$$
$$38\frac{6}{11}:2=38:2+\frac{6}{11}:2=19+\frac{6}{11}\cdot\frac{1}{2}=19+\frac{3}{11}=19\frac{3}{11}$$
$$100\frac{50}{109}:25=100:25+\frac{50}{109}:25=4+\frac{50}{109}\cdot\frac{1}{25}=4+\frac{2}{109}=4\frac{2}{109}$$
г)
$$49\frac{14}{15}:7=49:7+\frac{14}{15}:7=7+\frac{14}{15}\cdot\frac{1}{7}=7+\frac{2}{15}=7\frac{2}{15}$$
$$84\frac{42}{271}:42=84:42+\frac{42}{271}:42=2+\frac{42}{271}\cdot\frac{1}{42}=2+\frac{1}{271}=2\frac{1}{271}$$
$$100\frac{75}{647}:25=100:25+\frac{75}{647}:25=4+\frac{75}{647}\cdot\frac{1}{25}=4+\frac{3}{647}=4\frac{3}{647}$$
Ответ: а) $$2\frac{1}{9}$$; $$17\frac{2}{9}$$; $$7\frac{1}{8}$$; б) $$40\frac{2}{7}$$; $$2\frac{1}{3}$$; $$13\frac{4}{11}$$; в) $$16\frac{4}{21}$$; $$19\frac{3}{11}$$; $$4\frac{2}{109}$$; г) $$7\frac{2}{15}$$; $$2\frac{1}{271}$$; $$4\frac{3}{647}$$.
