Question

已知xyz=1，x+y+z=2，x²+y²+z²=16。 求1/xy+2z+1/yz+2x+1/zx+2y的值

Answer 1

z=2-x-y，所以1/(xy+2z)=1/(xy+4-2x-2y)=1/(x-2)(y-2)
设r=x-2,s=y-2,t=z-2
题目变为求1/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

Answer 2

(x+y+z)/xyz=1/xy+1/xz+1/yz=2
2x+2y+2z=4
1/xy+2z+1/yz+2x+1/zx+2y=2+4=6
..............无语啊。。。。