15 целых 3 28 ых - (4 целых 11 28 ых + y)=7-1целая - вопрос №3552277153284112871928 от sanekakot 04.08.2020 22:41

Question

Математика: 15 целых 3 28 ых - (4 целых 11 28 ых + y)=7-1целая 9 28 ых

sanekakot · Answer

5/7+3 6/7=3+(5+6)/7=3 11/7=4 4/7 1 6/11-7/11=17/11-7/11=(17-7)/11=10/11 9/15+7/15=(9+7)/15=16/15=1 1/15 9-3 7/8=5 8/8-7/8=5+(8-7)/8=5 1/8 6 6/13 -5 11/13=1 6/13-11/13=(19-11)/13=8/13 2 15/19+1 5/19=3 20/19=4 1/19 1 5/7-6/7=12/7-6/7=6/7 19/21+9/21=28/21=1 7/21 9 1/7-2 5/7= 7 1/7-5/7=6 3/7 6 8/9+ 7 5/9=13 13/9=14 4/9 6/7+5/7=11/7=1 4/7 5 3/8-3=2 3/8 7 9/17+8 8/17=15 17/17=16 9 3/13-8 9/13=1 3/13-9/13=(16-9)/13=7/13 1 6/11-7/11=(17-7)/11=10/11...