1)(7 целых 3/5-2 целых 4/15)*9/32 2)(11/24+1/6)*1 - вопрос №832411217352415932211241613535122 от azzitop 07.09.2022 07:18

Question

Математика: 1)(7 целых 3/5-2 целых 4/15)*9/32 2)(11/24+1/6)*1 целых 3/5 3)(5 целых 1/2+2 целых 3/5)*5/9 4)(7-2 целых 5/16)*4/25

azzitop · Answer

1) (7 3/5-2 4/15)×9/32=1 1/2
7 3/5-2 4/15= 5 (3/5-4/15)=5 (9-4/15)=
=5 1/3
5 1/3×9/32=16/3×9/32=3/2=1 1/2

2) (11/24+1/6)×1 3/5=1
11/24+1/6=11+4/24=15/24
15/24×1 3/5=15/24×8/5=3/3=1

3) (5 1/2+2 3/5)×5/9=4 1/2
5 1/2 +2 3/5=7 (1/2+3/5)=7 (5+6/10)=
=7 11/10=8 1/10
8 1/10×5/9=81/10×5/9=9/2 =4 1/2

4) (7-2 5/16)×4/25=3/4
7-2 5/16=6 16/16-2 5/16=4 11/16
4 11/16×4/25=75/16 ×4/25=3/4