Re: [解題] 一題多項式請高手幫忙！
※ 引述《YHank (Hank--since 2002/10)》之銘言：
: ※ 引述《sunfin (遠方)》之銘言：
: : 已知(x+y+z)(x+y)(y+z)(z+x)不等於零
: : 且[x^2/(y+z)]+[y^2/(z+x)]+[z^2/(x+y)]=0
: : 求:
: : 1. [x/(y+z)] + [y/(z+x)] + [z/(x+y)] = ?
: : 2. [x^2/(yz)] + [y^2/(zx)] + [z^2/(xy)] = ?
: : 感激不盡！
: 1. 設答案為w，
: w(x+y+z) = ......(請自行整理)
: = [x^2/(y+z)]+[y^2/(z+x)]+[z^2/(x+y)]+x+y+z = x+y+z
: (x+y+z)不為0，w=1
: 2. 已知[x/(y+z)]+[y/(z+x)]+[z/(x+y)] = 1
: 展開之，整理後得x^3+y^3+z^3+xyz=0
: 又xyz不為0，同除xyz得[x^2/(yz)]+[y^2/(zx)]+[z^2/(xy)]+1=0
: 所求為-1
: 其實我會想po是想請問第二小題是否有更elegant的解法，總覺得這樣解稍嫌暴力...
Here it is!
(x+y+z)(x+y)(y+z)(z+x) =/= 0
如果其中之一如x = 0,
y =/= -z 且 yz =/= 0
y^2/z + z^2/y = (y^3 + z^3)/yz = 0 yz 不為有限數
=> xyz =/= 0
[x^2/(y+z) + y^2/(x+z) + z^2/(x+y)][1/x + 1/y + 1/z] = 0
[x/(y+1) + y/(x+z) + z/(x+y)] + [x^2/(yz) + y^2/(xz) + z^2/(xy)] = 0
=> [x^2/(yz) + y^2/(xz) + z^2/(xy)] = -[x/(y+z) + y/(x+z) + z/(x+y)]
=> x^2/(yz) + y^2/(xz) + z^2/(xy) = -1
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