Cho x+y+z=0. Tính : A= (xy+2z^2)(yz+2x^2)(zx+2y^2)/(2xy^2+2yz^2+2zx^2+3xyz)^2

Có x+y+z=0

=> x+y=-z, y+z= -x và x+z =-y

Có x+y+z =0 

=> 3xyz= x³+ y³+z³

Xét tử của A

(xy+ 2z²)(yz+ 2x²)(zx+ 2y²)

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

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

= (xy+ z²- zx- zy)(yz+ x²- xy- xz)(zx+ y²- xy- yz)

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

= - (y-z)²(x-z)²(x-y)²

= - [(y-z)(x-z)(x-y)]²

= -( xyz- x²y- xz²+ x²z- y²z+ xy²+ yz²- xyz)²

= -( -x²y- xz²+ x²z- y²z+ xy²+ yz²)²

= -[x²(z-y)+y²(x-z)+ z²(y-x)]²

Xét mẫu của A

(2xy²+ 2yz²+ 2zx²+ 3xyz)²

= ( 2xy²+ 2yz²+ 2zx²+x³+ y³+ z³)²

= [ x²(2z+ x)+ y²( 2x+ y)+ z²( 2y+z)]²

= [ x²(z+ z+ x)+ y²( x+ x+ y)+ z²( y+ y+z)]²

= [ x²( z-y)+ y²(x- z)+ z²(y- x)]²

Có 

A= $\frac{(xy+ 2z²)(yz+ 2x²)(zx+ 2y²)}{(2xy²+ 2yz²+ 2zx²+ 3xyz)²}$ 

= $\frac{-[x²(z-y)+y²(x-z)+ z²(y-x)]²}{[ x²( z-y)+ y²(x- z)+ z²(y- x)]²}$ 

= -1