(3 1/3×1,3-7,2×2/27-9,2÷3,5)÷2/5 решить - вопрос №62055263131372227923525 от dimasergeenko 21.11.2021 07:54

Question

Алгебра: (3 1/3×1,3-7,2×2/27-9,2÷3,5)÷2/5 решить

dimasergeenko · Answer

(3 1/3 * 1,3 - 7,2 * 2/27 - 9,2 : 3,5) : 2/5 = 2 13/14

1) 3 1/3 * 1,3 = 10/3 * 13/10 = 13/3 = 4 1/3
2) 7,2 * 2/27 = 72/10 * 2/27 = (8*1)/(5*3) = 8/15
3) 9,2 : 3,5 = 92/10 * 10/35 = 92/35 = 2 22/35
4) 4 1/3 - 8/15 = 4 5/15 - 8/15 = 3 20/15 - 8/15 = 3 12/15 = 3 4/5
5) 3 4/5 - 2 22/35 = 3 28/35 - 2 22/35 = 1 6/35
6) 1 6/35 : 2/5 = 41/35 * 5/2 = (41*1)/(7*2) = 41/14 = 2 13/14