• Matéria: Matemática
  • Autor: uuhgtyuiytyui
  • Perguntado 3 anos atrás

a) ( x³ + 2x² + x ) : (+x) =
b) (x² + x³ + x⁴) : (+x²) =
c) (3x⁴ - 6x³ + 10x²) : (-2x²) =
d) (x⁷ + x⁵ + x³) : (-x²) =
e) (3x²y – 18xy²) : (+3xy) =
f) (7x³y – 8x²y²) : (-2xy) =
g) (4x²y + 2xy – 6xy²) : (-2xy) =
h) (20x¹² - 16x⁸ - 8x⁵) : ( +4x⁴) =
i) (3xy⁴ + 9x²y – 12xy²) : (+3xy) =


Mim ajudem aí pfvr!!!!!!! ​

Respostas

respondido por: lucasjoga450
3

a) \frac{\left(x^3+2x^2+x\right)}{\left(+x\right)}:\quad \left(x+1\right)^2

b) \frac{\left(x^2+x^3+x^4\right)}{\left(+x^2\right)}:\quad 1+x+x^2

c) \frac{\left(3x^4-6x^3+10x^2\right)}{\left(-2x^2\right)}:\quad -\frac{3x^2-6x+10}{2}

d) \frac{\left(x^7+x^5+x^3\right)}{\left(-x^2\right)}:\quad -x\left(x^2+1+x\right)\left(x^2+1-x\right)

e) \frac{\left(3x^2y-18xy^2\right)}{\left(+3xy\right)}:\quad x-6y

f) \frac{\left(7x^3y-8x^2y^2\right)}{\left(-2xy\right)}:\quad -\frac{x\left(7x-8y\right)}{2}

g) \frac{\left(4x^2y+2xy-6xy^2\right)}{\left(-2xy\right)}:\quad -2x-1+3y

h) \frac{\left(20x^{12}-16x^8-8x^5\right)}{\left(+4x^4\right)}:\quad x\left(5x^7-4x^3-2\right)

i) \frac{\left(3xy^4+9x^2y-12xy^2\right)}{\left(+3xy\right)}:\quad y^3+3x-4y

(Espero te ajudado, Bons estudos e tenha uma ótima noite.)

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