Question

1) Scrieti sub forma de produs:
a) 1502×2+1502×9=
b) 311+311×9=
c) 3468×104-2×3468-2×3468=
d) 2003×a-2×2003×b= ( a si b sunt numere naturale nenule, a>b).
2)Stiind ca a-c=37 si ab-bc+13=36b+42, calculati b si a+3b-c.
3) Calculati 11a+11b,stiind ca 8a+8b=64.
4) Corectati greseala, daca exista :
a) 6+120×6+80×6=6(120+80)
b) 2004+2003×2004+1×2003+1= 2004(1+2003+1)
Va rog mult ajutati-ma am nevoie, dau coroana!

Answer 1

1) Scrieti sub forma de produs:

a) 1502×2+1502×9=1502×(2 +9 ) = 1502 × 11

b) 311 + 311× 9 = 311×(1 +9) = 311 × 10

c) 3468×104-2×3468-2×3468= 3468 × (104 - 2 -2 ) = 3468 × 100

d) 2003×a-2×2003×b= 2003 × ( a - 2b)

( a si b sunt numere naturale nenule, a>b).

2)Stiind ca a - c=37 si ab - bc + 13= 36b + 42, calculati b si a + 3b - c

ab - bc + 13 = 36b + 42

Dam factor comun b(a - c)+ 13 = 36b+42

Dar a - c = 37 ⇒ 37b + 13 = 36b + 42 ⇒ b =29

Iar a + 3b - c = (a - c) + 3b = 37 + 3×29 = 37 + 87 = 124

3) Calculati 11a + 11b,stiind ca 8a + 8b = 64

8a + 8b = 64⇒ 8× (a + b) = 64⇒a + b = 64:8⇒a + b = 8

deci:

11a + 11b = 11× (a+b) = 11 × 8 = 88

4) Corectati greseala, daca exista :

a) 6+120×6+80×6 = 6×( 1+ 120+80) se da factor comun 6 intr-adevar dar mai apare un 1

nu poate fi egal cu 6(120+80)

b) 2004+2003×2004+1×2003+1= 2004(1+2003+1) + 1

se mai adauga un 1