Matemática, perguntado por isabelle7111, 9 meses atrás

me ajudem por favor!!

a) (x + 5).(6x²-12x +10)

b) (x-5).(x²+9x+10) =

c) (x + 7)(x²-x-2) =

d) (x + 5).(6x²-12x +10) =

e) (x + 5) (6x²-12x +10)=

f) (2x+5).(x²- 12x +10) =

g) (x + 5).(x²-12x +10) =

h) (5x-5).(x²- 12x +10) =​

Soluções para a tarefa

Respondido por Usuário anônimo
8

Explicação passo-a-passo:

a)

(x+5)\cdot(6x^2-12x+10)=6x^3-12x^2+10x+30x^2-60x+50

(x+5)\cdot(6x^2-12x+10)=6x^3+18x^2-50x+50

b)

(x-5)\cdot(x^2+9x+10)=x^3+9x^2+10x-5x^2-45x-50

(x-5)\cdot(x^2+9x+10)=x^3+4x^2-35x-50

c)

(x+7)\cdot(x^2-x-2)=x^3-x^2-2x+7x^2-7x-14

(x+7)\cdot(x^2-x-2)=x^3+6x^2-9x-14

d)

(x+5)\cdot(6x^2-12x+10)=6x^3-12x^2+10x+30x^2-60x+50

(x+5)\cdot(6x^2-12x+10)=6x^3+18x^2-50x+50

e)

(x+5)\cdot(6x^2-12x+10)=6x^3-12x^2+10x+30x^2-60x+50

(x+5)\cdot(6x^2-12x+10)=6x^3+18x^2-50x+50

f)

(2x+5)\cdot(x^2-12x+10)=2x^3-24x^2+20x+5x^2-60x+50

(2x+5)\cdot(x^2-12x+10)=2x^3-19x^2-40x+50

g)

(x+5)\cdot(x^2-12x+10)=x^3-12x^2+10x+5x^2-60x+50

(x+5)\cdot(x^2-12x+10)=x^3+7x^2-50x+50

h)

(5x-5)\cdot(x^2-12x+10)=5x^3-60x^2+50x-5x^2+60x-50

(5x-5)\cdot(x^2-12x+10)=5x^3-65x^2+110x-50


isabelle7111: man, vc salvou minha vida. obrigado!✨
Respondido por Makaveli1996
4

Oie, Td Bom?!

a)

(x + 5) \: . \: (6x {}^{2}  - 12x + 10) \\ x \: . \: (6x {}^{2}  - 12x + 10) + 5(6x {}^{2}  - 12x + 10) \\ 6x {}^{3}  - 12x {}^{2}  + 10x + 30x {}^{2}  - 60x + 50 \\ 6x {}^{3}  + 18x {}^{2}  - 50x + 50

b)

(x - 5) \: . \: (x {}^{2}  + 9x + 10) \\ x \: . \: (x {}^{2}  + 9x + 10) - 5(x {}^{2}  + 9x + 10) \\ x {}^{3}  + 9x {}^{2}  + 10x - 5x {}^{2}  - 45x - 50 \\ x {}^{3}  + 4x {}^{2}  - 35x - 50

c)

(x + 7) \: . \: (x {}^{2}  - x - 2) \\ x \: . \: (x {}^{2}  - x - 2) + 7(x {}^{2}  - x - 2) \\ x {}^{3}  - x {}^{2}  - 2x + 7x {}^{2}  - 7x - 14 \\ x {}^{3}  + 6x {}^{2}  - 9x - 14

d)

(x + 5) \: . \: (6x {}^{2}  - 12x + 10) \\ x \: . \: (6x {}^{2}  - 12x + 10) + 5(6x {}^{2}  - 12x + 10) \\ 6x {}^{3}  - 12x {}^{2}  + 10x + 30x {}^{2}  - 60x + 50 \\ 6x {}^{3}  + 18x {}^{2}  - 50x + 50

e)

(x + 5) \: . \: (6x {}^{2}  - 12x + 10) \\ x \: . \: (6x {}^{2}  - 12x + 10) + 5(6x {}^{2}  - 12x + 10) \\ 6x {}^{3}  - 12x {}^{2}  + 10x + 30x {}^{2}  - 60x + 50 \\ 6x {}^{3}  + 18x {}^{2}  - 50x + 50

f)

(2x + 5) \: . \: (x {}^{2}  - 12x + 10) \\ 2x \: . \: (x {}^{2}  - 12x + 10) + 5(x {}^{2}  - 12x + 10) \\ 2x {}^{3}  - 24x {}^{2}  + 20x + 5x {}^{2}  - 60x + 50 \\ 2x {}^{3}  - 19x {}^{2}  - 40x + 50

g)

( x+ 5) \: . \: (x {}^{2}  - 12x + 10) \\ x \: . \: (x {}^{2}  - 12x + 10) + 5(x {}^{2}  - 12x + 10) \\ x {}^{3}  - 12x {}^{2}  + 10x + 5x {}^{2}  - 60x + 50 \\ x {}^{3}  - 7x {}^{2}  - 50x + 50

h)

(5x - 5) \: . \: (x {}^{2}  - 12x + 10) \\ 5x \: . \: (x {}^{2}  - 12x + 10) - 5(x {}^{2}  - 12x + 10) \\ 5x {}^{3}  - 60x {}^{2}  + 50x + 5x {}^{2}  + 60x - 50 \\ 5x {}^{3}  - 65x {}^{2}  - 110x - 50

Att. Makaveli1996


Usuário anônimo: na última é -5.(x² - 12x + 10)
Usuário anônimo: tá sim
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