Question

QUESTÃO 01 Efetue as divisões: a) ( 12x² -8x) : (+2x) b) (3y³ + 6y²) : (3y) c) ( 10x² + 6x) : (-2x) d) (4x³ -9x) : (+3x) e) ( 15x³ -10x²) : (5x²) f) (30x² -20xy) : (-10x) g) (-18x² + 8x) : (+2x) h) (6x²y –4xy²) : (-2x) QUESTÃO 02 Efetue as Divisões: 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) QUEM PUDER ME AJUDAR, EU PRECISO DEMAIS DISSO, POR FAVOR

Answer 1

Explicação passo-a-passo:

1)

a)

$\sf \dfrac{12x^2-8x}{+2x}=6x-4$

b)

$\sf \dfrac{3y^3+6y^2}{3y}=y^2+2y$

c)

$\sf \dfrac{10x^2+6x}{-2x}=-5x-3$

d)

$\sf \dfrac{4x^3-9x}{+3x}=\dfrac{4x^2}{3}-3$

e)

$\sf \dfrac{15x^3-10x^2}{5x^2}=3x-2$

f)

$\sf \dfrac{30x^2-20xy}{-10x}=-3x+2y$

g)

$\sf \dfrac{-18x^2+8x}{+2x}=-9x+4$

h)

$\sf \dfrac{6x^2y-4xy^2}{-2x}=-3xy+2y^2$

2)

a)

$\sf \dfrac{x^3+2x^2+x}{+x}=x^2+2x+1$

b)

$\sf \dfrac{x^2+x^3+x^4}{+x^2}=1+x+x^2$

c)

$\sf \dfrac{3x^4-6x^3+10x^2}{-2x^2}=\dfrac{-3x^2}{2}+3x-5$

d)

$\sf \dfrac{x^7+x^5+x^3}{-x^2}=-x^5-x^3-x$

e)

$\sf \dfrac{3x^2y-18xy^2}{+3xy}=x-6y$

f)

$\sf \dfrac{7x^3y-8x^2y^2}{-2xy}=\dfrac{-7x^2}{2}+4xy$

g)

$\sf \dfrac{4x^2y+2xy-6xy^2}{-2xy}=-2x-1+3y$

h)

$\sf \dfrac{20x^{12}-16x^{8}-8x^5}{+4x^4}=5x^{8}-4x^4-2x$

i)

$\sf \dfrac{3xy^4+9x^2y-12xy^2}{+3xy}=y^3+3x-4y$

Answer 2

Oie, Td Bom?!

1.

a)

$\frac{12x {}^{2} - 8x}{2x} = \frac{2x \: . \: (6x - 4)}{2x} = 6x - 4$

b)

$\frac{3y {}^{ 3 } + 6y {}^{2} }{3y} = \frac{3y \: . \: (y {}^{2} + 2y)}{3y} = y {}^{2} + 2y$

c)

$\frac{10x {}^{2} + 6x}{ - 2x} = \frac{2x \: . \: (5x + 3)}{ - 2x} = - 5x - 3$

d)

$\frac{4x {}^{3} - 9x }{3x} = \frac{x \: . \: (4x {}^{2} - 9) }{3x} = \frac{4x {}^{2} - 9 }{3}$

e)

$\frac{15x {}^{3} - 10x {}^{2} }{5x {}^{2} } = \frac{5x {}^{2} \: . \: (3x - 2)}{5x {}^{2} } = 3x - 2$

f)

$\frac{30x {}^{2} - 20xy }{ - 10x} = \frac{10x \: . \: (3x - 2y)}{ - 10x} = - 3x + 2y$

g)

$\frac{18x {}^{2} + 8x}{2x} = \frac{2x \: . \: ( - 9x + 4)}{2x} = - 9x + 4$

h)

$\frac{6x {}^{2}y - 4xy {}^{2} }{ - 2x} = \frac{2x \: . \: (3xy - 2y {}^{2} )}{ - 2x} = - 3xy + 2y {}^{2}$

2.

a)

$\frac{x {}^{3} + 2x {}^{2} + x}{x} = \frac{x \: . \: (x {}^{2} + 2x + 1)}{x} = x {}^{2} + 2x + 1$

b)

$\frac{x {}^{2} + x {}^{3} + x {}^{4} }{x {}^{2} } = \frac{x {}^{2} \: . \: (1 + x + x {}^{2}) }{2} = x {}^{2} + x + 1$

c)

$\frac{3x {}^{4} - 6x {}^{3} + 10x {}^{2} }{ - 2x {}^{2} } = \frac{x {}^{2} \: . \: (3x {}^{2} - 6x + 10)}{ - 2x {}^{2} } = - \frac{3x {}^{2} - 6x + 10}{2}$

d)

$\frac{x {}^{7} + x {}^{5} + x {}^{3} }{ - x {}^{2} } = \frac{x {}^{2} \: . \: (x {}^{5} + x {}^{3} + x)}{ - x {}^{2} } = - x {}^{5} - x {}^{3} - x$

e)

$\frac{3x {}^{2}y - 18xy {}^{2} }{3xy} = \frac{3xy \: . \: (x - 6y)}{3xy} = x - 6y$

f)

$\frac{7x {}^{3}y - 8x {}^{2}y {}^{2} }{ - 2xy} = \frac{xy \: . \: (7x {}^{2} - 8xy)}{ - 2xy} = \frac{8xy - 7x {}^{2} }{2}$

g)

$\frac{4x {}^{2}y + 2xy - 6xy {}^{2} }{ - 2xy} = \frac{2xy \: . \: (2x + 1 - 3y)}{ - 2xy} = - 2x - 1 + 3y$

h)

$\frac{20x {}^{12} - 6x {}^{8} - 8x {}^{5} }{4x {}^{4} } = \frac{2x {}^{4} \: . \: (10x {}^{8} - 3x {}^{4} - 4x) }{4x {}^{4} } = \frac{10x {}^{8} - 3x {}^{4} - 4x}{2}$

i)

$\frac{3xy {}^{4} + 9x {}^{2}y - 12xy {}^{2} }{3xy} = \frac{3xy \: . \: (y {}^{3} + 3x - 4y)}{3xy} = y {}^{3} + 3x - 4y$

Att. Makaveli1996