Найдите среднее арифметическое чисел:
1) $\frac{5}{6}$ и $\frac{7}{20}$;
2) $1\frac{3}{7}$ и $2\frac{5}{21}$;
3) $2\frac{3}{5}, 3\frac{3}{10}$ и $2\frac{1}{2}$;
4) $7\frac{5}{24}, 6\frac{7}{24}$ и $8\frac{1}{6}$.
$(\frac{5}{6} + \frac{7}{20}) : 2 = (\frac{50}{60} + \frac{21}{60}) : 2 = \frac{71}{60} * \frac{1}{2} = \frac{71}{120}$
$(1\frac{3}{7} + 2\frac{5}{21}) : 2 = (1\frac{3}{7} + 2\frac{5}{21}) : 2 = (1\frac{9}{21} + 2\frac{5}{21}) * \frac{1}{2} = 3\frac{2}{3} * \frac{1}{2} = \frac{11}{3} * \frac{1}{2} = \frac{11}{6} = 1\frac{5}{6}$
$(2\frac{3}{5} + 3\frac{3}{10} + 2\frac{1}{2}) : 3 = (2\frac{6}{10} + 3\frac{3}{10} + 2\frac{5}{10}) : 3 = 8\frac{2}{5} : 3 = \frac{42}{5} * \frac{1}{3} = \frac{14}{5} = 2\frac{4}{5}$
$(7\frac{5}{24} + 6\frac{7}{24} + 8\frac{1}{6}) : 3 = (7\frac{5}{24} + 6\frac{7}{24} + 8\frac{4}{24}) : 3 = 21\frac{2}{3} * \frac{1}{3} = \frac{65}{3} * \frac{1}{3} = \frac{65}{9} = 7\frac{2}{9}$
Пожауйста, оцените решение
