Question

a+b+c=0求1/(a²+b²-c²)+1/(a²+c²-b²)+1/(b²+c²-a²)

帮忙了，聪明的你

Answer 1

a+b+c=0 a+b=-c a²+b²+2ab=c² a²+b²-c²=-2ab
同样方法可以推出 a²+c²-b²=-2ac b²+c²-a²=-2bc

1/(a²+b²-c²)+1/(a²+c²-b²)+1/(b²+c²-a²)=-1/2ab-1/2ac-1/2bc
=-1/2[(c+b+a)/abc]=0

Answer 2

解取a=1，b=-2，c=1代入1/(a²+b²-c²)+1/(a²+c²-b²)+1/(b²+c²-a²)
=1/4+1/(-2)+1/(4)=1/2-1/2=0

Answer 3

a+b+c=0 c=-(a+b) C^2=(a+b)^2 a^2+b^2-c^2=a^2+b^2-(a+b)^2=-2ab 同理a^2+c^2-b^2=-2ac
b^2+c^2-a^2=-2bc 所以最后化简得1/-2ab+1/-2ac+1/-2bc=(a+b+c)/-2abc=0