(7 3/52 4/15)*9/32 (11/24+1/6)*1 3/5 (5 1/2+2 3/5)*5/9 - вопрос №489053273524159321124161355122355972516425 от serbinaluba66cet 16.05.2020 13:04

Question

Математика: (7 3/52 4/15)*9/32 (11/24+1/6)*1 3/5 (5 1/2+2 3/5)*5/9 (7-2 5/16)*4/25

serbinaluba66cet · Answer

(1 1/2 + 2/3)х6 = (3/2 + 2/3)х6 = (9/6 + 4/6)х6 = 13/6х6=13 ; (2 1/2 + 1 1/3) : 1/6 - 2 3/7 =( 5/2 +4/3) :1/6 -  2 3/7 = (15/6 + 8/6) :1/6 -  2 3/7 == 23/6  :1/6 -  2 3/7 = 23 -  2 3/7 = 20 4/7  (1 3/5 + 2 1/5): 1 9/10 + (1 7/10 - 4/5): 3/7 = 3  4/5 :1 9/10  + (17/10 - 8/10) : 3/7 == 19/5 Х 10/19 + 9/10 Х 7/3 = 2 + 2 1/10 = 4 1/10 ;   1 8/9:(3 1/6-1 3/4)=  1 8/9:(19/6 - 7/4) =   1 8/9: 38-21/12 =    1 8/9:17/12 == 17/9 Х  12/17 = 12/9 = 4/3 = 1 1/3...