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15 tháng 10 2021

\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)

15 tháng 10 2021

phân tích ra  ii chj làm vậy ai chả lm đc

6 tháng 10 2018

      \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

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

15 tháng 10 2016

a) \(x^4-2x^3+2x-1\)

\(=x^4-x^3-x^3+2x-2+1\)

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

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

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

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

\(=\left(x-1\right)\left[\left(x^3-x^2\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)

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

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

\(=\left(x-1\right)^3\left(x+1\right)\)

b) \(x^4+2x^3+2x^2+2x+1\)

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

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

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

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

8 tháng 8 2018

\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)

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

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

\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)

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

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

\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)

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

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

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

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

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

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

8 tháng 8 2018

\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)

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

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

\(x^3-4x^2+12x-27\)

\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)

\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)

\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)

\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)

\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)

29 tháng 1 2019

\(x^4+2x^3+3x^2+2x+1.\)

\(=x^4+x^3+x^3+x^2+x^2+x^2+x+x+1\)
\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)

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

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

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

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

\(=\left(x+1\right)^4\)

29 tháng 1 2019

@wi

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

\(x^2+2x+1=\left(x+1\right)^2\)chứ

25 tháng 8 2021

\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)

\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)

\(=x^8+12x^5-2x^4+36x^2-12x-99\)

\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)

\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)

\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)

16 tháng 10 2020

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

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

= (x - 5 + 2x - 1)2 - (2x - 6)2

= (3x - 6)2 - (2x - 6)2

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

16 tháng 10 2020

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

= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2

= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2

= ( 3x - 6 )2 - ( 2x - 6 )2

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

= x( 5x - 12 )

25 tháng 10 2017

Ta có : \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^3+x^2+x^2+x+1\)

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

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