№3 a)\(\frac{5}{9}+\frac{10}{27}\); 6)\(\frac{7}{10}-\frac{3}{20}\); в)\(\frac{6}{25}\cdot\frac{5}{18}\); г)\(5\frac{2}{7}-2\frac{4}{21}+3\frac{2}{3}\)

Задание 3

1. \(\frac{5}{9}+\frac{10}{27}=\frac{5\cdot3}{9\cdot3}+\frac{10}{27}=\frac{15}{27}+\frac{10}{27}=\frac{25}{27}\)
2. \(\frac{7}{10}-\frac{3}{20}=\frac{7\cdot2}{10\cdot2}-\frac{3}{20}=\frac{14}{20}-\frac{3}{20}=\frac{11}{20}\)
3. \(\frac{6}{25}\cdot\frac{5}{18}=\frac{6\cdot5}{25\cdot18}=\frac{30}{450}=\frac{1}{15}\)
4. \(5\frac{2}{7}-2\frac{4}{21}+3\frac{2}{3} = \frac{37}{7}-\frac{46}{21}+\frac{11}{3} = \frac{37\cdot3}{7\cdot3}-\frac{46}{21}+\frac{11\cdot7}{3\cdot7} = \frac{111}{21}-\frac{46}{21}+\frac{77}{21} = \frac{111-46+77}{21} = \frac{142}{21} = 6\frac{16}{21}\)