Question

va rog ajutatima

Dacă x = 5 și a + b = 13, calculați:

a) 3 ⋅ x + 7 ⋅ a + 7 ⋅ b
b) x ⋅ a + x ⋅ b– 50
c) 10 ⋅ x + (4 ⋅ a + 4 ⋅ b)
d) (4 ⋅ a + 4 ⋅ b– 2 ⋅ x) ⋅ (2 ⋅ a + 2 ⋅ b + x).

Answer 1

$x = 5$

$a + b = 13$

$a)3x + 7a + 7b = 3x + 7(a + b)$

$= 3 \times 5 + 7 \times 13$

$= 15 + 91$

$= 106$

$b)xa + xb - 50 = x(a + b) - 50$

$= 5 \times 13 - 50$

$= 65 - 50$

$= 15$

$c)10x + 4a + 4b = 10x + 4(a + b)$

$= 10 \times 5 + 4 \times 13$

$= 50 + 52$

$= 102$

$d)(4a + 4b - 2x)(2a + 2b + x)$

$= [4(a + b) - 2x][2(a + b) + x]$

$= (4 \times 13 - 2 \times 5)(2 \times 13 + 5)$

$= (52 - 10)(26 + 5)$

$= 42 \times 31$

$= 1302$

Answer 2

x = 5

a + b = 13

__________

a. 3 ⋅ x + 7 ⋅ a + 7 ⋅ b  

b. x ⋅ a + x ⋅ b– 50  

c. 10 ⋅ x + (4 ⋅ a + 4 ⋅ b)

d. (4 ⋅ a + 4 ⋅ b– 2 ⋅ x) ⋅ (2 ⋅ a + 2 ⋅ b + x)

a. 3 ⋅ x + 7 ⋅ a + 7 ⋅ b =

pas ₁: Se dă factor comun pe 7.

3 ⋅ x + 7 ⋅ (a + b ) =

pas ₂: Se înlocuieşte.

3 ⋅ 5 + 7 ⋅ 13=

pas₃: Se rezolvă.

15+ 91=

106

b. x ⋅ a + x ⋅ b– 50=

pas ₁: Se dă factor comun pe x.

x ⋅ (a + b )- 50=

pas ₂: Se înlocuieşte.

5·13- 50=

pas₃: Se rezolvă.

65-50=

15

c. 10 ⋅ x + (4 ⋅ a + 4 ⋅ b) =

pas ₁: Se dă factor comun pe 4.

10 ⋅ x + 4 ⋅ (a+ b) =

pas ₂: Se înlocuieşte.

10 ⋅ 5 + 4 ⋅ 13=

pas₃: Se rezolvă.

50+ 52=

102

d. (4 ⋅ a + 4 ⋅ b– 2 ⋅ x) ⋅ (2 ⋅ a + 2 ⋅ b + x) =

pas ₁: Se dă factor comun pe 4 şi 2.

[4 ⋅ (a + b– 2 ⋅ x)] ⋅ [2 ⋅ (a + b) + x] =

pas ₂: Se înlocuieşte.

(4 ⋅ 13 -2·5)·(2⋅ 13+ 5)=

pas₃: Se rezolvă.

(52-10)·(26+ 5)=

42·31=

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