x+y+z=0,则1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2=多少?
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x+y+z=0,则1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2=多少?
x+y+z=0,则1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2=多少?
x+y+z=0,则1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2=多少?
把a,b,c换成x,y,z就行了：
a+b+c=0
a+b=-c
(a+b)^2=c^2
a^2+b^2-c^2=-2ab
同理
b^2+c^2-a^2=-2cb
a^2+c^2-b^2=-2ac
所以原式=-1/2ab-1/2bc-1/2ac=-(a+b+c)/abc=0
参考资料：baidu
已知x+y+z=0，则1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2)=-
x=-y-z 所以x^2=(y+z)^2
所以y^2+z^2-x^2=-2yz y^2=(x+z)^2
所以z^2+x^2-y^2=-2xz z^2=(x+y)^2
所以x^2+y^2-z^2=-2xy
所以原式 =1/(-2yz)+1/(-2xz)+1/(-2xy)
=x/(-2xyz)+y/(-2xyz)+z/(-2xyz)
=(x+y+z)/(-2xyz)
=0
x+y+z=0
y+z=-x
两边平方
y^2+2yz+z^2=x^2
y^2+z^2-x^2=-2yz
同理
z^2+x^2-y^2=-2zx
x^2+y^2-z^2=-2xy
所以1/(y^2+z^2-x^2)+1/(z^2+x^2-y^2)+1/(x^2+y^2-z^2)
=-1/2yz-1/2zx-1/2xy
=-(1/2)(1/yz+1/zx+1/xy)
=-(1/2)[(x+y+z)/xyz]
=0