已知a^2+b^2+c^2=1,a*(1/b 1/c)+b*(1/a 1/c)+c*(1/a 1/b)=0,求a=?,b=?,c=?

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=(a+b)/c+(b+c)/a+(c+a)/b
=c/c+(a+b)/c+a/a+(b+c)/a+b/b+(c+a)/b-3
=(a+b+c)(1/a+1/b+1/c)-3 ＝0
(a+b+c)(1/a+1/b+1/c)＝3

给你个参考题目：
a^2+b^2+c^2=1,a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=(a+b)/c+(b+c)/a+(c+a)/b
=c/c+(a+b)/c+a/a+(b+c)/a+b/b+(c+a)/b-3
=(a+b+c)(1/a+1/b+1/c)-3
则(a+b+c)(1/a+1/b+1/c)=0
a+b+c=0,or ab+bc+ca=0
当ab+bc+ca=0
则(a+b+c)²=a²+b²+c²+2ab+2bc+2ca=1
则a+b+c=±1
故a+b+c=0,or ±1
故a+b+c=0,or ±1
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