Question

REZOLVATI IN 2 MODURI
4×5±5=
6±3×7=
21×3±4=
5±4×10

Answer 1

Răspuns:

Explicație pas cu pas:

4 × ( 5 + 5 ) = 4 × 10 = 40 → primul mod

4 × 5 + 4 × 5 = 20 + 20 = 40 → al doilea mod ( distributivitate)

4 × ( 5 - 5 ) = 4 × 0 = 0 → primul mod

4 × 5 - 4 × 5 = 20 - 20 = 0 → al doilea mod

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( 6 + 3 ) × 7 = 9 × 7 = 63 → primul mod

6 × 7 + 3 × 7 = 42 + 21 = 63 → al doilea mod

( 6 - 3 ) × 7 = 3 × 7 = 21 → primul mod

6 × 7 - 3 × 7 = 42 - 21 = 21 → al doilea mod

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21 × ( 3 + 4 ) = 21 × 7 = 147 → primul mod

21 × 3 + 21 × 4 = 63 + 84 = 147 → al doilea mod

21 × ( 4 - 3 ) = 21 × 1 = 21 → primul mod

21 × 4 - 21 × 3 =84 - 63 = 21 → al doilea mod

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21 × ( 3 - 4 ) = 21 × ( - 1 ) = - 21

21 × 3 - 21 × 4 = 63 - 84 = - 21

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( 5 + 4 ) × 10 = 9 × 10 = 90 → primul mod

5 × 10 + 4 × 10 = 50 + 40 = 90 → al doilea mod

( 5 - 4 ) × 10 = 1 × 10 = 10 → primul mod

5 × 10 - 4 × 10 = 50 - 40 = 10 → al doilea mod