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14 tháng 12 2020

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

 

25 tháng 2 2021

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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

 

25 tháng 8 2016

a) 5x2 ( 3x2 -7x+2)-15x(x-3)

=15x4-35x3+10x2-15x2+45x

=15x4-35x3-5x2+45x

c) (x+3)(x-3)(x-2)(x+1)

=(x2-9)(x2+x-2x-2)

=(x2-9)(x2-x-2)

=x4-x3-2x2-9x2+9x+18

=x4-x3-11x2+9x+18

d)(2x+1)2+(4x-1)2+2(2x+1)(4x+1)

=2x2+4x+1-16x2-8x+1

=2x2+4x+1-16x2-8x+1+16x2-4x+8x-2

=2x2+7

e) (2x2-3x)(5x2-2x+1)-10x2(x+3)

=10x4 -4x3+2x2-15x3+6x2-3 -10x2-30x

=10x4-19x3-2x2-30x-3

26 tháng 8 2016

thanks bn nka

a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)

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

\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)

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

\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

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

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

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

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

 

15 tháng 9 2023

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)