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已知xyz=1,x+y+z=2,x^2+y^2+z^2=16.求1/9xy+2z)+1/(yz+2x)+1/(zx+2y)的值
人气:330 ℃ 时间:2019-08-20 21:06:22
解答
xy + xz + yz = ((x+y+z)^2 - (x^2+y^2+z^2))/2 = -6x^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...
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