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若x+y+z=7,x^2+y^2+z^2=11,xyz=6,求：1/x+1/y+1/z=?
题目详情
若x+y+z=7,x^2+y^2+z^2=11,xyz=6,求：1/x+1/y+1/z=?
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答案和解析
x+y+z=7
=>
(x+y+z)=49
=>
x^2 + y^2 + z^2 + 2xy + 2yz + 2xz = 49
=>
xy + yz + xz = (49-11)/2 = 19
1/x+1/y+1/z = (xy + yz + xz)/xyz = 19/6
=>
(x+y+z)=49
=>
x^2 + y^2 + z^2 + 2xy + 2yz + 2xz = 49
=>
xy + yz + xz = (49-11)/2 = 19
1/x+1/y+1/z = (xy + yz + xz)/xyz = 19/6
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