2个回答
展开全部
(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)
=[(x-y)+(x-z)]/(x-y)(x-z)+[(y-z)+(y-x)]/(y-z)(y-x)+[(z-x)+(z-y)]/(z-x)(z-y)
=1/(x-z)+1/(x-y)+1/(y-x)+1/(y-z)+1/(z-y)+1/(z-x)
=1/(x-z)+1/(x-y)-1/(x-y)+1/(y-z)-1/(y-z)-1/(x-z)
=0
=[(x-y)+(x-z)]/(x-y)(x-z)+[(y-z)+(y-x)]/(y-z)(y-x)+[(z-x)+(z-y)]/(z-x)(z-y)
=1/(x-z)+1/(x-y)+1/(y-x)+1/(y-z)+1/(z-y)+1/(z-x)
=1/(x-z)+1/(x-y)-1/(x-y)+1/(y-z)-1/(y-z)-1/(x-z)
=0
展开全部
(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)
=[(y-z)(2x-y-z)-(x-z)(2y-z-x)+(x-y)(2z-x-y)]/[(x-y)(x-z)(y-z)]
=[2xy-2xz-y²+z²-2xy+2yz-x²+z²+2xz-2yz-x²+y²)/
=2(z²-x²)/[(x-y)(x-z)(y-z)]
=-2(x+z)/[(x-y)(y-z)]
=[(y-z)(2x-y-z)-(x-z)(2y-z-x)+(x-y)(2z-x-y)]/[(x-y)(x-z)(y-z)]
=[2xy-2xz-y²+z²-2xy+2yz-x²+z²+2xz-2yz-x²+y²)/
=2(z²-x²)/[(x-y)(x-z)(y-z)]
=-2(x+z)/[(x-y)(y-z)]
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