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)=?

Answer 1

x+y+z=2平方
x^2+y^2+z^2+2xy+2xz+2yz=4
x^2+y^2+z^2=16
2xy+2xz+2yz=4-16=-12
1/(xy+2x)=1/(xy+4-2x-2y)=1/(x-2)/(y-2)
所以原式=1/(x-2)/(y-2)+1/(y-2)/(z-2)+1/(z-2)/(x-2)
=(x+y+z+6)/(xyz-2xy-2yz-xz+4x+4y+4z-8)=-4/13

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Answer 2

把后面的式子通分，然后化简 ，带入已知条件，就可以了

Answer 3

=(x+y+z+6)/(xzy-2xy-3zy-zx+4x+4y+4z-8)=-4/13

Answer 4

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)
= -52/169

参考资料：http://zhidao.baidu.com/question/49975457.html?si=2