急急急!又一道高中数学题!谁来帮
由(1/a)+(1/b)+(1/c) = 1/(a+b+c)得:
[(1/a)+(1/b)+(1/c)](a+b+c)=1,
[(a+b)/ab+(1/c)][(a+b)+c]=1,
(a+b)^2/ab+c(a+b)/ab+(a+b)/c+1=1,
(a+b)[(a+b)/ab+c/ab+1/c]=0,
故a+b=0或(a+b)/ab+c/ab+1/c=0,
由a+b=0得a=-b,
由(a+b)/ab+c/ab+1/c=0得
(a+b+c)/ab=-1/c,
ac+bc+c^2+ab=0,
(b+c)(a+c)=0,
故b+c=0或a+c=0,即b=-c或a=-c,
所以由(1/a)+...全部
由(1/a)+(1/b)+(1/c) = 1/(a+b+c)得:
[(1/a)+(1/b)+(1/c)](a+b+c)=1,
[(a+b)/ab+(1/c)][(a+b)+c]=1,
(a+b)^2/ab+c(a+b)/ab+(a+b)/c+1=1,
(a+b)[(a+b)/ab+c/ab+1/c]=0,
故a+b=0或(a+b)/ab+c/ab+1/c=0,
由a+b=0得a=-b,
由(a+b)/ab+c/ab+1/c=0得
(a+b+c)/ab=-1/c,
ac+bc+c^2+ab=0,
(b+c)(a+c)=0,
故b+c=0或a+c=0,即b=-c或a=-c,
所以由(1/a)+(1/b)+(1/c) = 1/(a+b+c)可得出:
a=-b或b=-c或a=-c,
因为n 是奇数,
所以,当a=-b时,
[1/(a^n)]+[1/(b^n)]+[1/(c^n)]
=[1/(-b)^n]+[1/(b^n)]+[1/(c^n)]
=-[1/(b^n)]+[1/(b^n)]+[1/(c^n)]
=1/(c^n),
1/[(a^n)+(b^n)+(c^n)]
=1/[(-b)^n+(b^n)+(c^n)]
=1/(-b^n+b^n+c^n)
=1/(c^n),
1/[(a+b+c)^n]
=1/[(-b+b+c)^n]
=1/(c^n),
所以[1/(a^n)]+[1/(b^n)]+[1/(c^n)]=1/[(a^n)+(b^n)+(c^n)]
=1/[(a+b+c)^n]=1/(c^n),
同理,当b=-c或a=-c时,也可证得
[1/(a^n)]+[1/(b^n)]+[1/(c^n)]=1/[(a^n)+(b^n)+(c^n)]=1/[(a+b+c)^n],
综上所述,若(1/a)+(1/b)+(1/c) = 1/(a+b+c) 及 n 是奇数,则
[1/(a^n)]+[1/(b^n)]+[1/(c^n)]=1/[(a^n)+(b^n)+(c^n)]=1/[(a+b+c)^n]。
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