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13 tháng 10 2017

x 2 y + x y 2  +  x 2 z + x z 2  +  y 2 z + y z 2  + 3xyz.

= ( x 2  y +  x 2 z + xyz) + (x y 2  +  y 2 z + xyz) + (x z 2  + y z 2  + xyz)

= x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)

= (x + y + z)(xy + xz + yz).

14 tháng 12 2020

\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

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

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

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

25 tháng 12

x2(y - z) + y2(z - x) + z2(x - y)

 

= z2(x - y) + x2 y - x2 z + y2 z - y2 x

 

= z2(x - y) + (x2 y - y2 x) + (- x2 z + y2 z)

 

= (x - y)(z2 + xy - zx - zy)

 

= (x - y)[(z2 - zx) + (xy - zy)]

 

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

27 tháng 7 2018

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

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

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

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

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

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

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

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

26 tháng 7 2017

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

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

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

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

chúc bn hc tốt ^^ 

6 tháng 8 2017

a,Từ giả thiết ta có

(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2

=(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2

Đặt x2+y2+z2=a

xy+yz+zx=b

=>(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2

=a(a+2b)+b2

=a2+2ab+b2

=(a+b)2

=(x2+y2+z2+xy+yz+zx)2

câu b hơi dài mình gửi sau nhé

6 tháng 8 2017

Ta có: 2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4

Gọi x^4+y^4+z^4=a

x^2+y^2+z^2=b

x+y+z=c

=>2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4=2a-b^2-2bc^2+c^4

=2a-2b^2+b^2-2bc^2+c^4

=2(a-b^2)+(b+c^2)^2

Ta có

2(a-b2)=2[x^4+y^4+z^4-(x^2+y^2+z^2)2]

=2[x^4+y^4+z^4-x^4-y^4-z^4-2x2y2-2y2z2-2z2x2]

=2.(-2)(x2y2+y2z2+z2x2)

=-4(x2y2+y2z2+z2x2)

Lại có

(b+c^2)^2

=[(x^2+y^2+z^2)+(x+y+z)2]2

=[(x^2+y^2+z^2)-(x^2+y^2+z^2)-2(xy+yz+zx)]2

=4(xy+yz+zx)2

=>2(a-b^2)+(b+c^2)^2

=-4(x2y2+y2z2+z2x2)+4(xy+yz+zx)2

=8xyz(x+y+z)

11 tháng 10 2020

Ta có: \(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)

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

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

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

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

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

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

\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

17 tháng 11 2019

a) (x - 1)(x + l)(x - 2)(x - 4).      b) (x - 2)( x 2  + 4).

c) 2y(3 x 2   +   y 2 ).                          d) 2(x + y + z) ( a   -   b ) 2 .

24 tháng 8 2021

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

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

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

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

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

b. \(x^3-2x^2+4x-8\)

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

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

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

c. \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=6x^2y+2y^3\)

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

d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)

\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)

\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)

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

27 tháng 10 2019

Câu hỏi của nguyễn khánh linh - Toán lớp 8 - Học toán với OnlineMath

10 tháng 11 2021

x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz

=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz

=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz

=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3

=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]

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

=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]

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

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