Re: [中學] 一題數學
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03/12 19:05,
03/12 19:05
來回答這個問題
a式: x + y + z = w
b式: 1/x + 1/y + 1/z = 1/w
a代入b: 1/x + 1/y + 1/z = 1/(x+y+z)
(xy+yz+yz)/xyz = 1/(x+y+z)
由於 x,y,z,x+y+z都不為0
所以：
(x+y+z)(xy+yz+zx) = xyz
=> (x+y+z)(xy+yz) + (x+y+z)*xz = xyz
=> y(x+y+z)(x+z) + xz(x+z) = 0
=> (x+z)[y(x+y+z)+xz] = 0
=> (x+z)[y^2+(x+z)y+xz] = 0
=> (x+z)(y+x)(y+z) = 0
換言之， x =-z or x =-y or y =-z
即： w = y or w = z or w = x
所以沒有4個數都相異的解
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※ 編輯: ckchi 來自: 140.116.127.158 (03/12 20:41)
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