Question

a+b+c=0,求证1/(b^2+c^2-a^2)+1/(c^2+a^2-b^2)+1/(a^2+b^2-c^2)=0

Answer 1

解：因为a+b+c=0
所以a=-b-c
b=-a-c
c=-a-b
a^2=(b+c)^2=b^2+2bc+c^2
b^2=(a+c)^2=a^2+2ac+c^2
c^2=(a+b)^2=a^2+2ab+b^2
代入就可以得到：
1/(b^2+c^2-a^2)+1/(c^2+a^2-b^2)+1/(a^2+b^2-c^2)=0
=1/(b^2+c^2-b^2-2bc-c^2)+1/(c^2+a^2-a^2-2ac-c^2)+1/(a^2+b^2-a^2-2ab-b^2)
=-1/2[1/(bc)+1/(ac)+1/(ab)]
=-1/2[(a+b+c)/abc]
=0

以上希望对你的理解有帮助~谢谢~~~~

Answer 2

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

希望我的回答对你有帮助，采纳吧O(∩_∩)O！

Answer 3

a+b+c=0,故a^2=(b+c)^2 ; b^2=(a+c)^2 ; c^2=(a+b)^2

三个分母：b^2+c^2-a^2=b^2+c^2-(b+c)^2= -2bc
c^2+a^2-b^2= -2ac
a^2+b^2-c^2= -2ab
原式= -1/2bc-1/2ac-1/2ab= -(a+b)/2abc-c/2abc= -(a+b+c)/2abc = 0

Answer 4

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