Question

Resolva os monômios abaixo se possível:

a. (6ax/2) + 5bx – (12 ax/3) – (10bx/5) + 4x=
b. [15y . 4z . (3xy)² ]: (30y . 9z) =

Answer 1

Resposta:

Explicação passo a passo:

a.

(6ax/2) + 5bx – (12 ax/3) – (10bx/5) + 4x=

= 3ax + 5bx - 4ax - 2bx + 4x

= - ax + 3bx + 4x

= x.(-a + 3b + 4)

b. [15y . 4z . (3xy)² ]: (30y . 9z) =

15y/30y = 1/2

4z/9z = 4/9

= 9x^2.y^2 . 1/2 . 4/9

= 9/9 . x^2.y^2 . 4/2

= 2.x^2.y^2

Answer 2

Resposta:

$\left(\frac{6ax}{2}\right)+5bx-\left(\frac{12ax}{3}\right)-\left(\frac{10bx}{5}+4x\right)\\\\\frac{6ax}{2}+5bx-\frac{12ax}{3}-\left(\frac{10bx}{5}+4x\right)\\\\3ax+5bx-\frac{12ax}{3}-\left(\frac{10bx}{5}+4x\right)\\\\3ax+5bx-4ax-\left(\frac{10bx}{5}+4x\right)\\\\3ax+5bx-4ax-\left(2bx+4x\right)\\\\-\left(2bx+4x\right)=-2bx-4x\\\\3ax+5bx-4ax-2bx-4x\\\\=-ax+3bx-4x$

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$\frac{15y\cdot \:4z\left(3xy\right)^2}{30y\cdot \:9z}\\\\\frac{15\cdot \:4z\left(3xy\right)^2}{30\cdot \:9z}\\\\\frac{15\cdot \:4\left(3xy\right)^2}{30\cdot \:9}\\\\\frac{15\cdot \:4\left(3xy\right)^2}{15\cdot \:2\cdot \:9}\\\\\frac{4\left(3xy\right)^2}{2\cdot \:9}\\\\\frac{2\cdot \:2\left(3xy\right)^2}{2\cdot \:9}\\\\\frac{2\left(3xy\right)^2}{9}\\\\\frac{2\cdot \:3^2x^2y^2}{9}\\\\\frac{2\cdot \:3^2x^2y^2}{3^2}\\\\=2x^2y^2$