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

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)

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

b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)

\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)

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

a: \(=x^2-4-x^2+x+8=x+4\)

a) (x-2)(x+2)-x(x-1)+8
= x²-4-x²+x+8
= (x²-x²)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

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

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

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

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,

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

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

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

b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)

=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)

=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)

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

=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)

=\(\dfrac{x+3}{x}\)

#Fiona

c) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)

=\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{3^2-x^2}\) + \(\dfrac{x}{x+3}\)

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

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

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

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

=\(\dfrac{3^2+2.3x+x^2}{\left(x-3\right)\left(x+3\right)}\)

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

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

=\(\dfrac{x-3}{x+3}\)

#Fiona 

Tick đúng giúp mình nhaa<3

d)\(\dfrac{5x+10}{4x-8}\).\(\dfrac{x-2}{x+2}\)

=\(\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\) . \(\dfrac{x-2}{x+2}\)

=\(\dfrac{5\left(x+2\right).\left(x-2\right)\text{​​}\text{​​}}{4\left(x-2\right).\left(x+2\right)}\)

=\(\dfrac{5}{4}\)

#Fiona

Tick đúng giúp mikk nhaa

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Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)