已知abc≠0，且a+b+c=o，求a*(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值

显示全部楼层 · 2009-6-4 20:12:29

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千问 · 2009-6-4 20:12:29

a+b+c=0a+b=-c b+c=-a a+c=-b a*(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=(a/b0+(a/c)+(b/c)+(b/a)+(c/a)+(c/b)=(a/b)+(c/b)+(b/a)+(c/a)+(a/c)+(b/c)=(a+c/b)+(c+b/c)+(a+b/c)将a+b=-c b+c=-a a+c=-b带如式子中得 (-b/b)+(-a/a)+(-c/c)=(-1)+(-1)+(-1)=-3...

千问 · 2009-6-4 20:12:29

a+b+c=0 b+c=-a,a+b=-c,a+c=-ba*(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=abc/(b+c)+abc/(a+c)+abc/(a+b)=abc(1/(b+c)+1/(a+c)+1/(a+b))=abc(-a-b-c)=-abc(a+b+c)=0...