• Matéria: Matemática
  • Autor: thaisslopcucn8
  • Perguntado 8 anos atrás

como se resolve [5/7-2/21] + [3/4-1/7]

Respostas

respondido por: Anônimo
0

= [5/7-2/21] + [3/4-1/7]
= [(3.5 - 1.2)/21] + [(7.3 - 4.1)/28]
= [(15 - 2)/21] + [(21 - 4)/28]
= 13/21 + 17/28
= (4.13 + 3.17)/84
= (52 + 51)/84
= 103/84

21,28: 2
21,14: 2
21,7: 3
7,7: 7
1,1 = 2.2.3.7 = 4.21 = 84

R.: 103/84

respondido por: marinaldoferrepcqb0f
0

mmc(7,21) = 21 e mmc(4,7) = 28

[5/7 - 2/21] + [3/4 - 1/7]

[15/21 - 2/21] + [21/28 - 4/28]

13/21 + 17/28

mmc(21,28) = 84

52/84 + 51/84 = 103/84

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