已知a+b+c=0,abc不等于0,求1/（b*+c*-a*）+1除以(c*+a*-b*）+1除以（a*+b*-c*)=
问题描述：
已知a+b+c=0,abc不等于0,求1/（b*+c*-a*）+1除以(c*+a*-b*）+1除以（a*+b*-c*)=
答
由於a+b+c=0
即
a=-(b+c) or b+c=-a
b=-(a+c) or a+c=-b
c=-(a+b) or a+b=-c
1/（b*+c*-a*）+1除以(c*+a*-b*）+1除以（a*+b*-c*)
= 1/(-a-a)+1/(-b-b)+1/(-c-c)
= 1/(-2a)+1/(-2b)+1/(-2c)
= (-1/2)x(1/a+1/b+1/c)
= (-1/2)x(bc+ca+ab)/(abc)
= (-1/2)x(bc+a(c+b))/(abc)
= (-1/2)x(bc-(c+b)^2)/(abc)
= (1/2)x(c^2+cb+b^2)/(abc)
= (1/2)x(a^2-cb)/(abc)
答
∵a+b+c=0
∶∴a=-(b+c)
∴b²+c²-a²=b²+c²-(b²+2bc+c²)=-2bc
同理c²+a²-b²=-2ca
a²+b²-c²=-2ab
∴原式=-1/2bc-1/ca-1/2ab
=-(a+b+c)/2abc
=0