已知xyz=1,x+y+z=2,x²+y²+z²=16. 求1/xy+2z+1/yz+2x+1/zx+2y的值
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![已知xyz=1,x+y+z=2,x²+y²+z²=16. 求1/xy+2z+1/yz+2x+1/zx+2y的值](/uploads/image/z/568072-64-2.jpg?t=%E5%B7%B2%E7%9F%A5xyz%3D1%2Cx%2By%2Bz%3D2%2Cx%26%23178%3B%2By%26%23178%3B%2Bz%26%23178%3B%3D16.+%E6%B1%821%2Fxy%2B2z%2B1%2Fyz%2B2x%2B1%2Fzx%2B2y%E7%9A%84%E5%80%BC)
已知xyz=1,x+y+z=2,x²+y²+z²=16. 求1/xy+2z+1/yz+2x+1/zx+2y的值
已知xyz=1,x+y+z=2,x²+y²+z²=16. 求1/xy+2z+1/yz+2x+1/zx+2y的值
已知xyz=1,x+y+z=2,x²+y²+z²=16. 求1/xy+2z+1/yz+2x+1/zx+2y的值
(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.
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