a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)

b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)

c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)

b,

\(=xy+2xy^2-4\)

c,

\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)

\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

\(a,=2xy^3\\ b,=xy+3xy^2-4\\ c,=\left(x-4\right)\left(x+4\right):\left(x-4\right)=x+4\\ d,=\left(x+5\right)^2:\left(x+5\right)=x+5\)

(15³y²-6x²y-3x²y²):6x²y

=\(\dfrac{5}{2}\)xy-1-\(\dfrac{1}{2}\)y

(4x²-9y²):(2x-3y)

=(2x-3y)(2x+3y):(2x-3y)

=2x+3y

\(b,=3xy\left(xy+2xy^2-4\right):3xy\\ =xy+2xy^2-4\\ c,=\left[\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)\right]:\left(2x^2-x+1\right)\\ =\left[x\left(2x^2-x+1\right)+3\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\\ =\left[\left(x+3\right)\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\\ =x+3\)

b: \(=xy+2xy^2-4\)

D

TỪ BẠN ƠI

a: =9x^2-12x+4-4x^2+14x

=5x^2+2x+4

b: \(=\dfrac{2+x+1+x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x+2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)

Vậy (6x3 – x2 – 14x + 3) : (2x – 3) = 3x2 + 4x – 1

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)