已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,求

xy+2z=xy+4-2x-2y=(x-2)(y-2).
yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).
4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),
xy+yz+zx=-6.
(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.
原式=[(x-2)+(y-2)+(z-2)]/(x-2)(y-2)(z-2)
=-4/13.

z=2-x-y，所以1/(xy+2z)=1/(xy+4-2x-2y)=1/(x-2)(y-2)
設(shè)r=x-2,s=y-2,t=z-2
題目變?yōu)榍?/rs+1/st+1/rt=(r+s+t)/rst
而由已知可得(r+2)(s+2)(t+2)=1，r+s+t=-4，(r+2)^2+(s+2)^2+(t+2)^2=16
解得(r+s+t)/rst=-4/13