Question

已知a,b,c为非零实数，1/a+1/b+1/c不等于0，a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=—3 求a+b+c

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Answer 1

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=—3
a/b+a/c+b/c+b/a+c/a+c/b+3=0
a/b+c/b+1+a/c+b/c+1+b/a+c/a+1=0
(a+b+c)/b+(a+b+c)/c+(a+b+c)/a=0
(a+b+c)的平方/abc=0
a+b+c=0

追问

过程？

追答

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=—3
a/b+a/c+b/c+b/a+c/a+c/b+3=0
a/b+c/b+1+a/c+b/c+1+b/a+c/a+1=0
(a+b+c)/b+(a+b+c)/c+(a+b+c)/a=0
(a+b+c)/(1/b+1/c+1/a)=0
因为1/a+1/b+1/c不等于0
所以a+b+c=0

Answer 2

令a+b+c=t
则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=（t-b）/b+（t-a）/a+（t-c）/c
=t（1/b+1/c+1/a）-3=-3
即t（1/b+1/c+1/a）=0
因为1/a+1/b+1/c不等于0，所以t=0
即a+b+c=0