Question

已知a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值

Answer 1

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

Answer 2

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=(a/b+c/b)+(a/c+b/c)+(b/a+c/a)=(a+c)/b+(a+b)/c+(b+c)/a
∵a+b+c=0
∴b=-c-a c=-b-a a=-b-c
∴原式=(-b)/b+(-c)/c+(-a)/a=-1-1-1=-3.