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答：
x+y+z=0
y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1
=1/(y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)
=1/[y²+z²-(-y-z)²]+1/[z²+x²-(-z-x)²]+1/[x²+y²-(-x-y)²]
=1/(-2yz)+1/(-2xz)+1/(-2xy)
=-(1/2)*(x+y+z)/(xyz)
=0
x+y+z=0
y²+z²-x²分之1+z²+x²-y²分之1+x²+y²-z²分之1
=1/(y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)
=1/[y²+z²-(-y-z)²]+1/[z²+x²-(-z-x)²]+1/[x²+y²-(-x-y)²]
=1/(-2yz)+1/(-2xz)+1/(-2xy)
=-(1/2)*(x+y+z)/(xyz)
=0
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