Question

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calculeaza in 2 moduri
7x(4+6)
(9-5)x9
3x(5+2)

Answer 1

Primul mod :

7 × ( 4 + 6 ) =

7 × 10 = 70

Al doilea mod :

7 × 4 + 7 × 6 =

28 + 42 = 70

Primul mod :

( 9 - 5 ) × 9 =

4 × 9 = 36

Al doilea mod :

9 × 9 - 5 × 9 =

81 - 45 = 36

Primul mod :

3 × ( 5 + 2 ) =

3 × 7 = 21

Al doilea mod :

3 × 5 + 3 × 2 =

15 + 6 = 21

Answer 2

7×(4+6)=70

7×10=70

7×1×(4+6)=70

7×(4+6)

7×10=70

(9-5)×9=36

4×9=36

(9-5-0)×1×9=36

4×9=36

3×(5+2)=21

3×7=21

3×1×(5×1+2×1)=21

3×7=21