作业帮已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1/（xy+2z)+1/(yz+2x)+1/(zx+2y)=?
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已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1/（xy+2z)+1/(yz+2x)+1/(zx+2y)=?
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1/（xy+2z)+1/(yz+2x)+1/(zx+2y)=?
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1/（xy+2z)+1/(yz+2x)+1/(zx+2y)=?
xy + xz + yz = ((x+y+z)^2 - (x^2+y^2+z^2))/2 = -6
1/（xy+2z)=1/[xy+2(1-1/xy)]=x^2y^2+2/xy
通例1/(yz+2x)=y^2z^2+2/yz
1/(zx+2y)=x^2z^2+2/xz
通分即可,答案是-2
通分
z/(xyz+2z^2)+x/(xyz+2x^2)+y/(xyz+2y^2)
=z/(1+2z^2)+x/(1+2x^2)+y/(1+2y^2)
xy + xz + yz = ((x+y+z)^2 - (x^2+y^2+z^2))/2 = -6
x^2y^2 + x^2z^2 + y^2z^2 = (xy + xz + yz)^2 - 2xyz(x+y+z) = 32
原式 = ((yz+2x)(xz+2y) + (xy+2z)(xz+2y) + (xy+2z)(yz+2x)) / (xy+2z)(xz+2y)(yz+2x)...
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xy + xz + yz = ((x+y+z)^2 - (x^2+y^2+z^2))/2 = -6
x^2y^2 + x^2z^2 + y^2z^2 = (xy + xz + yz)^2 - 2xyz(x+y+z) = 32
原式 = ((yz+2x)(xz+2y) + (xy+2z)(xz+2y) + (xy+2z)(yz+2x)) / (xy+2z)(xz+2y)(yz+2x)
= (xyz^2 + 2x^2y + 2y^2z + 4xy + x^2yz + 2xy^2 + 2xz^2 + 4yz + xy^2z + 2x^2y + 2yz^2 + 4xy) / (x^2y^2z^2 + 2x^3yz + 2xy^3z + 2xyz^3 + 4x^2y^2 + 4x^2z^2 + 4y^2z^2 + 8xyz)
= (xyz(x+y+z) + 2(xy+xz+yz)(x+y+z) + 4(xy+xz+yz) - 6xyz)
/ ((xyz)^2 + 2xyz(x^2+y^2+z^2) + 4(x^2y^2 + x^2z^2 + y^2z^2) + 8xyz)
= -4/13
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