化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?

来源：学生作业帮助网编辑：作业帮时间：2024/04/29 23:12:25

化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?

化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?
令a = x - y,b = y - z,c = z - x
原式= (a - c)/(-ac) + (b - a)/(-ba) + (c - b)/(-cb)
= (a - b)/ab + (b - c)/bc + (c - a)/ca
= [(ac - bc) + (ab - ac) + (bc - ab)]/abc
= 0/abc
= 0

原式=（x-y)+(x-z)/(x-y)(x-z)+(y-z)+(y-x)/(y-z)(y-x)+(z-y)+(z-x)/(z-x)(z-y)
=(x-y)/(x-y)(x-z)+(x-z)/(x-y)(x-z)+(y-z)/(y-z)(y-x)+(y-x)/(y-z)(y-x)+(z-y)/(z-x)(z-y)+(z-x)/(z-x)(z-y) =1/(x-z）+1/(x-y)+1/(y-x)+1/(y-z)+1/(z-x）+1/(z-y)
=[1/(x-z）+/(z-x）]+[1/(x-y)+1/(y-x)]+[1/(y-z)+1/(z-y)]
=0

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