1.已知a+b+c=0,得:
a+c=-b
a+b=-c
b+c=-a
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
=(a+c)/b+(a+b)/c+(b+c)a
=(0-b)/b+(0-c)/c+(0-a)/a
=(-1)+(-1)+(-1)
=-3
2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/（c^2+a^2-b^2)+ab/(a^2+b^2-c^2)
已知a+b+c=0,得:
a+c=-b
a+b=-c
b+c=-a
bc/(b^2+c^2-a^2)+ac/（c^2+a^2-b^2)+ab/(a^2+b^2-c^2)
=bc/[b^2+c^2-(b+c)^2]+ac/[c^2+a^2-(a+c)^2]+ab/[a^2+b^2-(a+b)^2)]
=bc/(-2bc)+ac/(-2ac)+ab/(-2ab)
=(-1)/2+(-1)/2+(-1)/2
=-1.5