Question

设有理数a,b,c都不为0,且a+b+c=0,则1/(b²+c²-a²)+1/(c

Answer 1

a+b+c=0

a+b=-c

(a+b)^2=c^2

a^2+b^2-c^2=2ab

同理有：

a^2+c^2-b^2=2ac

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

所以：

1/（b^2+c^2-a^2）+1/(c^2+a^2-b^2)+1/（a^2+b^2-c^2）=1/(2bc)+1/(2ac)+1/2ab

=(a+b+c)/2abc

=0