Question

Calculeaza in doua moduri (2+3)×5; (4+5)×8; (9-4)×8; (8-3)×7; (2+1+3)×4; (5+1+2)×7

Answer 1

Răspuns:

5×5=2×5+3×5=25;  9×8=4×8+5×8=72 ;5×8=9×8-4×8=40 ;5×7=8×7-3×7=35;     6×4=2×4+1×4+3×4=24;  8×7=6×7+2×7=56

Explicație pas cu pas:

Answer 2

Răspuns:

(2+3)×5=

5×5=25

sau

(2+3)×5=

2×5+3×5=

10+15=25

(4+5)×8=

9×8=72

sau

(4+5)×8=

4×8+5×8=

32+40=72

(9-4)×8=

5×8=40

SAU

(9-4)×8=

9×8-4×8=

72-32=40

(8-3)×7=

5×7=35

sau

(8-3)×7=

8×7-3×7=

56-21=35

(2+1+3)×4=

(3+3)×4=

6×4=24

SAU

(2+1+3)×4=

2×4+1×4+3×4=

8+4+12=

12+12=24

(5+1+2)×7=

8×7=56

sau

(5+1+2)×7=

5×7+1×7+2×7=

35+7+14=

42+14=56