Question

calculeaza in doua moduri 6o.3supra 5 ; 5supra 6.9; 14. 3supra 4;

Answer 1

Răspuns:

Explicație pas cu pas:

1. a)    60 ×  3 / 5 = 180 / 5 = 36 → primul mod

          ( 60 / 5 ) ×  3 = 12 × 3 = 36 → al doilea mod

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b)     ( 5 / 6 ) × 9 = ( 5 × 9 ) / 6 = 45 / 6 =(45 :3 )/( 6:3) = 15/2 → primul mod

       5 × ( 9 / 6 ) = 5 × ( 3 / 2 ) = ( 5 × 3 ) / 2 = 15 / 2 → al doilea mod

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c)   14 × ( 3 / 4 ) =  ( 14 × 3 ) / 4 = 42 / 4 = ( 42:2)/(4:2) = 21 / 2

      ( 14 / 4 ) × 3 = ( 7 / 2 ) × 3 = ( 7 × 3) / 2 = 21 / 2

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2)

a)   7 intregi si 2 / 3 = ( 7 x 3 + 2 ) / 3 = 23 / 3

( 23 / 3 ) × 6 = ( 23 × 6 ) / 3 = 138 / 3 = 46 → primul mod

23 × ( 6 / 3 ) = 23 × 2 = 46 → al doilea mod

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b) 3 intregi si 2 / 5 = ( 3 × 5 + 2 ) / 5 = 17 / 5

= ( 17 / 5 ) × 15 = ( 17 × 15 ) / 5 = 255 / 5 = 51 → primul mod

= 17 × ( 15 / 5 ) = 17 × 3 = 51 → al doilea mod

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c) 4 intregi si 3 / 7 = ( 4 × 7 + 3 ) / 7 = 31 / 7

= ( 31 / 7 ) × 5 = ( 31 × 5 ) / 7 = 155 / 7 = 22 intregi si 1 / 7

= 31  × ( 5 / 7 ) = 155 / 7