Question

实数a,b,c都不为0,且a+b+c=0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值是几?

Answer 1

解:

a+b+c=0

所以

a+b=-c

a+c=-b

b+c=-a

所以

a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b)

=a/b+a/c+b/c+b/a+c/a+c/b

=(b+c)/a+(a+c)/b+(a+b)/c

=(-a)/a+(-b)/b+(-c)/c

=-1-1-1

=-3