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?

问题描述:

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?

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

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