Question

Найдите значения суммы и разности чисел (10-11): 11.1)7 целых 11/12и 4целых
9/16
25целых 13/42 и 19целых 9/14
5целых 7/30 и 3целых 11/18
49целых 3/20 и 38целых 1/12​

Answer 1

Пошаговое объяснение:

$1)7 \frac{11}{12} + 4 \frac{9}{16} = 7 \frac{44}{48} + 4 \frac{27}{48} = 11 \frac{44 + 27}{48} = 11 \frac{71}{48} = 12 \frac{23}{48} \\ 7 \frac{11}{12} - 4 \frac{9}{16} = 7 \frac{44}{48} - 4 \frac{27}{48} = 3 \frac{44 - 27}{48} = 3 \frac{17}{48} \\ 2)25 \frac{13}{42} + 19 \frac{9}{14} =25 \frac{13}{42} + 19 \frac{27}{42} = 44 \frac{13 + 19}{42} = 44 \frac{32}{42} = 44 \frac{16}{21} \\ 25 \frac{13}{42} - 19 \frac{9}{14} = 25 \frac{13}{42} - 19 \frac{27}{42} = 5 \frac{55 - 27}{42} = 5 \frac{28}{42} = 5 \frac{2}{3} \\ 3)5 \frac{7}{30} + 3 \frac{11}{18} = 5 \frac{21}{90} + 3 \frac{55}{90} = 8 \frac{76}{90} = 8 \frac{38}{45} \\ 5 \frac{7}{30} - 3 \frac{11}{18} = 5 \frac{21}{90} - 3 \frac{55}{90} = 1 \frac{111 - 55}{90} = 1 \frac{56}{90} = 1 \frac{28}{45} \\ 4)49 \frac{3}{20} + 38 \frac{1}{12} = 49 \frac{9}{60} + 38 \frac{5}{60} = 87 \frac{14}{60} = 87\frac{7}{30} \\ 49 \frac{3}{20} - 38 \frac{1}{12} = 49 \frac{9}{60} - 38 \frac{5}{60} = 11 \frac{4}{60} = 11 \frac{1}{15}$