若a,b,c都不为o,a+b+c=0,求1/(b^2+c^2-a^2)+1/(c^2+a^2-b^2)+1/(a^2+b^2-c^2)的值

b^2+c^2-a^2=(b+c)^2-2bc-a^2=a^2--2bc-a^2=-2bc

同理c^2+a^2-b^2=-2ac
a^2+b^2-c^2=-2ab

所以原式=1/（-2bc)+1/(-2ac)+1/(-2ab)
=(a+b+c)/(-2abc)
=0

a=-b-c,
1/[b^2+c^2-(-b-c)^2]=1/(2bc)
1/(c^2+a^2-b^2)=1/(2ac)
1/(a^2+b^2-c^2)=1/(2ab)
1/(2bc)+1/(2ac)+1/(2ab)
=1/4[2/(bc)+2/(ac)+2/(ab)]
1/(bc)+1/(ac)=(1/c)*(a+b)/(ab)=(a+b)/abc
1/4[2(a+b+c)/abc]=0
写的很草``我用公式编辑器写给你``
你等下

b^2+c^2-a^2=c^2+(b-a)(b+a)=c^2-c(b-a)=c(c-b+a)=-2bc
c^2+a^2-b^2=a^2+(c-b)(c+b)=a^2-a(c-b)=a(a-c+b)=-2ac
a^2+b^2-c^2=b^2+(a-c)(a+c)=b^2-b(a-c)=b(b-a+c)=-2ab
(-1/2)(1/bc+1/ca+1/ab)=(-1/2)[(1/b+1/a)/c+1/ab]
=(-1/2)[(a+b)/(abc)+1/ab]
=(-1/2)[(-c)/(abc)+1/ab]
=0

a+b=-c
(a+b)^2=c^2
a^2+b^2-c^2=-2ab
同理b^2+c^2-a^2=-2bc,a^2+c^2-b^2=-2ac
1/(b^2+c^2-a^2)+1/(c^2+a^2-b^2)+1/(a^2+b^2-c^2)
=-(1/(2ab)+1/(2bc)+1/(2ac))
=-(a+b+c)/(2abc)
=0