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)=?](/uploads/image/z/976388-68-8.jpg?t=%E5%B7%B2%E7%9F%A5xyz%3D1%2Cx%2By%2Bz%3D2%2Cx%5E2%2By%5E2%2Bz%5E2%3D16%2C%E5%88%991%2F%EF%BC%88xy%2B2z%29%2B1%2F%28yz%2B2x%29%2B1%2F%28zx%2B2y%29%3D%3F)
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