Lời giải:

a.

$2x(3x^2-4x+2)=2x.3x^2-2x.4x+2x.2$

$=6x^3-8x^2+4x$

b.

$2x(3x+5)-3(2x^2-2x+3)=2x.3x+2x.5-(3.2x^2-3.2x+3.3)$

$=6x^2+10x-6x^2+6x-9=16x-9$

=3x(x^2-2)(3x^2+x-2)

=(3x^3-6x)(3x^2+x-2)

=9x^5+3x^4-6x^3-18x^3-6x^2+12x

=9x^5+3x^4-12x^3-6x^2+12x

2x(x^2-1)=2x^3-2x

\(a,=12x^2+18x-12x^2+x+1=19x+1\)

\(a,=12x^2+18x-12x^2+x+1=19x+1\\ b,=4x^3y^4-3xy^2-\dfrac{3}{2}x\)

\(\frac{x}{3x+1}+\frac{2x+1}{3x+1}=\frac{x+2x+1}{3x+1}=\frac{3x+1}{3x+1}=1\)

\(\text{CHÚC BẠN HOK TOT}\)

\(\frac{x}{3x+1}+\frac{2x+1}{3x+1}=\frac{x+2x+1}{3x+1}=\frac{\left(x+2x\right)+1}{3x+1}=\frac{3x+1}{3x+1}=1\)

\(=\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}+\dfrac{2-3x}{\left(x+1\right)^2}\\ =\dfrac{\left(3x+2\right)\left(x+1\right)^2-6\left(x^2-1\right)+\left(2-3x\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\\ =\dfrac{10x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)

Lời giải:
a.

$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$

$=5x-(-4x+3)=5x+4x-3=9x-3$

b.

$(-3x^2+2x-1)+(4x^2-2x+3)$

$=-3x^2+2x-1+4x^2-2x+3=x^2+2$

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

=5x^2+2x+4

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