数学分式问题！

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a/b+a/c+b/a+b/c+c/a+c/b
=(b+c)/a+(a+c)/b+(a+b)/c
因为a+b+c=0
所以a+b=-c,b+c=-a,a+c=-b
(b+c)/a+(a+c)/b+(a+b)/c
=(-a)/a+(-b)/b+(-c)/c
=-1-1-1
=-3

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

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

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

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

=ac(a+c)/abc + ab(a+b)/abc + bc(b+c)/abc

=(a^2c+ac^2+a^2b+ab^2+b^2c+bc^2)/abc

=[a^2(c+b)+b^2(a+c)+c^2(a+b)]/abc