koglomik
Rozwiązane

1zad (4/10-1)×(3 1/3-2,5) pod spodem 1/3

2) oblicz

a)2,4:1 1/5=
b)2 4/7:4/21=
c)5/8×3/20=
d)12/25×5/9=
e)6/13×20/21×7/10=
3)
2/3:3=
2,4:0,4=
8/4:4/7=
720:0,8
4zad
2,4:1 1/5=
2 4/7:4/21=​

Odpowiedź :

Odpowiedź:

1)

[(4/10-1) × (3 1/3-2,5)] / 1/3 = [(4/10-10/10) × (10/3-25/10)] / 1/3 = [-6/10 × (100/30 - 75/30)] : 1/3 = -6/10 × 25/30 × 3 = -3/2 = -1 1/2 = -1,5

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2)

a) 2,4 : 1 1/5 = 24/10 : 6/5 = 24/10 × 5/6 = 4/2 = 2

b) 2 4/7 : 4/21 = 18/7 × 21/4 = 9*3 / 2 = 27/2 = 13 1/2

c) 5/8 × 3/20 = 3 / 8×4 = 3/32

d) 12/25 × 5/9 = 4 / 5×3 = 4/15  

e) 6/13 × 20/21 × 7/10 = 2×2 / 13 = 4/13

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3)

2/3 : 3 = 2/3 × 1/3 = 2/9

2,4 : 0,4 = 24 : 4 = 6 lub 24/10 × 10/4 = 24/4 = 6

8/4 : 4/7 = 8/4 × 7/4 = 2×7 / 4 = 7/2 = 3 1/2 = 3,5

720 : 0,8 = 7200 : 8 = 900

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4)

2,4 : 1 1/5 = 24/10 : 6/5 = 24/10 : 12/10 = 24/10 × 10/12 = 24/12 = 2

2 4/7 : 4/21 =​ 18/7 × 21/4 = 18×3 / 4 = 54/4 = 13 1/2

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