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4 tháng 12 2014

= x3 + y3 + z3 + 3x2yz + 3xy2z + 3xyz2 - x3 -y3 - z3

=3x2yz + 3xy2z + 3xyz2

= 3xyz( x + y + z)

4 tháng 12 2014

b.

x^4+2012x^2+2012x-x+2012=

(x^4-x)+2012(x^2+x+1)=

x(x-1)(x^2+x+1)+2012(x^2+x+1)=

(x+2012)(x^2+x+1)

 

1 tháng 9 2020

a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)^3+z^3+3.\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)

\(=\left[x^3+y^3+3xy.\left(x+y\right)+z^3+3\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)

\(=3xy\left(x+y\right)+3\left(x+y\right)z.\left(x+y+z\right)\)

\(=3.\left(x+y\right)\left(xy+zx+zy+z^2\right)\)

\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

b) \(x^4+2012x^2+2011x+2012\)

\(=x^4-x+2012x^2+2012x+2012\)

\(=x.\left(x^3-1\right)+2012.\left(x^2+x+1\right)\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)

9 tháng 9 2017

\(a\text{)}\left(x+y+z\right)^3-x^3-y^3-z^3\)

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

\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz-y^2+yz-z^2\right)\)

\(=\left(y+z\right)\left(3x^2+3xy+3yz+3xz\right)\)

\(=3\left(y+z\right)\left(x^2+xy+yz+xz\right)\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

\(b\text{)}x^4+2012x^2+2011x+2012\)

\(=\left(x^4-x\right)+\left(2012x^2+2012x+2012\right)\)

\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)

\(=\left(x^2-x\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)

\(=\left(x^2-x+2012\right)\left(x^2+x+1\right)\)

18 tháng 8 2019

x4+2012x2+2011x+2012

=(x4-x)+(2012x2+2012x+2012)

=x(x3-1)+2012(x2+x+1)

=x(x-1) (x2+x+1) + 2012 (x2+x+1)

=(x2+x+1) [x(x-1)+2012]

=(x2+x+1) (x2-x+2012)

1 tháng 9 2020

\(x^4+2012x^2+2011x+2012\)

\(=x^4-x+2012x^2+2012x+2012\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)

20 tháng 8 2017

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

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

2) \(x^4+2012x^2+2011x+2012\)

\(=\left(x^4-x\right)+\left(2012x^2+2012x+2012\right)\)

\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2012\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+2012\right]\)

\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)

11 tháng 7 2018

t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !

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

11 tháng 7 2018

\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)

\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)

\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)

\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)

\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)

Mk ko chắc

22 tháng 7 2017

a) \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-xz-yz+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

b) \(x^4+2011x^2+2010x+2011\)

\(=x^4+2010x^2+x^2+2010x+2010+1\)

\(=\left(x^4+x^2+1\right)+\left(2010x^2+2010x+2010\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2010\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)

5 tháng 9 2018

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

13 tháng 12 2020

a) (x  + y + z)3 - x3 - y3 - z3

= (x + y + z)3 - z3 - (x3 + y3

= (x + y + z - z)[(x + y + z)2 + (x + y + z).z + z2) - (x + y)(x2 - xy + y2)

= (x + y)(x2 + y2 + z2 + 2xy + 2yz + 2zx + 2xz + 2yz + z2 + z2) - (x + y)(x2 - xy + y2)

= (x + y)(x2 + y2 + 3z2 + 2xy + 4yz + 4zx) - (x + y)(x2 - xy + y2)

= (x + y)(3z2 + 3xy + 5yz + 4zx) 

b) Sửa đề x4 + 2010x2 + 2009x + 2010

= (x4 + x2 + 1) + (2009x2 + 2009x + 2009)

= (x4 + 2x2 + 1 - x2) + 2009(x2 + x + 1)

= [(x2 + 1)2 - x2] + 2009(x2 + x + 1)

= (x2 + x + 1)(x2 - x + 1) + 2009(x2 + x + 1)

= (x2 + x + 1)(x2 - x + 2010)

29 tháng 10 2017

a, x4 - 5x2 + 4

= x4 - 4x- x+ 4

= x2 . (x2 - 4) - (x2 - 4)

= (x2 - 4) . (x- 1)

= (x - 2) . (x + 2) . (x - 1) . (x + 1)

29 tháng 10 2017

a)x4-5x2+4=x4-x2-4x2+4
                   =(x4-x2)-(4x2-4)
                   =x2(x2-1)-4(x2-1)
                    =(x2-1)(x2-4)