Задача 69274 решить 5 и 6...
Условие
математика 6-7 класс
83
Решение
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[m]4\frac{2}{3}+(-2\frac{8}{15})-(-2\frac{3}{5})=4\frac{2}{3}-2\frac{8}{15}+2\frac{3}{5}=4\frac{10}{15}-2\frac{8}{15}+2\frac{9}{15}=4+\frac{10}{15}-2-\frac{8}{15}+2+\frac{9}{15}=[/m]
[m]=(4-2+2)+(\frac{10}{15}-\frac{8}{15}+\frac{9}{15})=4+\frac{11}{15}=4\frac{11}{15}[/m]
6
[m]4\frac{2}{3}-(-3\frac{3}{4}+(-2\frac{5}{6})-5\frac{1}{8}=4\frac{2}{3}+3\frac{3}{4}-2\frac{5}{6})-5\frac{1}{8}=[/m]
[m]=4+\frac{2}{3}+3+\frac{3}{4}-2-\frac{5}{6}-5-\frac{1}{8}=(4+3-2-5)+(\frac{2\cdot 8}{3\cdot 8}+\frac{3\cdot 6}{4\cdot 6}-\frac{5\cdot 4}{6\cdot 4}-\frac{1\cdot 3}{8\cdot 3})=0+(\frac{16}{24}+\frac{18}{24}-\frac{20}{24}-\frac{3}{24})=\frac{11}{24}[/m]