Question

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

Answer 1

a+b+c=0
a+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