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7 tháng 8 2020

a, (x4-2x3+2x-1):(x2-1) = \(\frac{\left(x^4-1\right)-\left(2x^3-2x\right)}{x^2-1}\) 

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

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

b, (8x3-6x2-5x+3):((4x+3) 

9 tháng 11 2017

Mn ơi làm hộ mk vs mai mk kiểm tra

5 tháng 9 2020

Mấy bài dài dài kia tí mình làm cho :) 

( x - 1 )3 - x( x - 2 )2 + 1 

= x3 - 3x2 + 3x - 1 - x( x2 - 4x + 4 ) + 1

= x3 - 3x2 + 3x - x3 + 4x2 - 4x

= x2 - x = x( x - 1 )

2x( 3x + 2 ) - 3x( 2x + 3 )

= 6x2 + 4x - 6x2 - 9x

= -5x

( x + 2 )3 + ( x - 3 )2 - x2( x + 5 )

= x3 + 6x2 + 12x + 8 + x2 - 6x + 9 - x3 - 5x2

= 2x2 + 6x + 17

( 2x + 3 )( x - 5 ) + 2x( 3 - x ) + x - 10

= 2x2 - 7x - 15 + 6x - 2x2 + x - 10

= -25

( x + 5 )( x2 - 5x + 25 ) - x( x - 4 )2 + 16x

= x3 + 53 - x( x2 - 8x + 16 ) + 16x

= x3 + 125 - x3 + 8x2 - 16x + 16

= 8x2 + 125

( -x - 2 )3 + ( 2x - 4 )( x2 + 2x + 4 ) - x2( x - 6 )

= -x3 - 6x2 - 12x - 8 + 2x3 - 16 - x3 + 6x2

= -12x - 24 = -12( x + 2 )

5 tháng 9 2020

Tương tự ... 

a, \(\left(x-1\right)^3-x\left(x-2\right)^2+1=x^3-3x^2+3x-1-x^3+4x^2-4x+1=x^2-x\)

b, \(2x\left(3x+2\right)-3x\left(2x+3\right)=6x^2+4x-6x^2-9x=-5x\)

c, \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)=x^3+6x^2+12x+8+x^2+6x+9-x^3-5x^2=2x^2+18x+17\)

11 tháng 10 2020

a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)

b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

c) \(2x^3-x^2-8x+4\)

\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)

d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)

\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-y^2+x\right)\)

e) \(2x^2-5x+2\)

\(=\left(2x^2-x\right)-\left(4x-2\right)\)

\(=x\left(2x-1\right)-2\left(2x-1\right)\)

\(=\left(x-2\right)\left(2x-1\right)\)

a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)

b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)

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

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

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

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

d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)

\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)

27 tháng 10 2020

a) ( 3x - 2 )( 4x + 5 ) - 6x( 2x - 1 )

= 12x2 + 7x - 10 - 12x2 + 6x

= 13x - 10

b) ( 2x - 5 )2 - 4( x + 3 )( x - 3 )

= 4x2 - 20x + 25 - 4( x2 - 9 )

= 4x2 - 20x + 25 - 4x2 + 36

= 61 - 20x

c) 2x3 - 5x2 + 7x - 6

= 2x3 - 3x2 - 2x2 + 3x + 4x - 6

= x2( 2x - 3 ) - x( 2x - 3 ) + 2( 2x - 3 )

= ( 2x - 3 )( x2 - x + 2 )

=> ( 2x3 - 5x2 + 7x - 6 ) : ( 2x - 3 ) = x2 - x + 2

27 tháng 10 2020

a, (3x - 2 ) (4x + 5) - 6x (2x -1) = ( 7x + 15x -8x - 10 ) - ( 12x2 -6x ) = 7x2 + 15x - 8x -10 -12x2 + 6x = -5x2 + x - 10 

a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

\(=\left(6x^2+x-2\right)\left(3-x\right)\)

\(=18x^2-6x^3+3x-x^2-6+2x\)

\(=-6x^3+17x^2+5x-6\)

c) Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)\)

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)

\(=x^3+1\)

e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)

\(=10x^4+4x^3-15x^2-6x-5x-2\)

\(=10x^4+4x^3-15x^2-11x-2\)

f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)

\(=x^3-4x^2-2x^2+8x+3x-12\)

\(=x^3-6x^2+11x-12\)

g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)

\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)

\(=7x^2+16x-1-x^2+7x-12\)

\(=6x^2+23x-23\)

h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)

\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)

\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)

\(=-5x^3+26x^2-28x-2\)