∵x-2y+z=(x-y)-(y-z),x+y-2z=(y-z)-(z-x),y+z-2x=(z-x)-(x-y).
设x-y=a,y-z=b,z-x=c,则
原式=-ac/ (a-b)(b-c) +(-ba)/(b-c)(c-a) +(-cb)/(c-a)(a-b)
=-ac(c-a)+ba(a-b)+bc(b-c)/(a-b)(b-c)(c-a)
=-ac²-a²c+ba²-b²a+b²c-bc² /(a-b)(b-c)(c-a)
=-c²(a-b)-c(a²-b²)+ab(a-b) /(a-b)(b-c)(c-a)
=-(a-b)(c²-ca-cd+ab)/(a-b)(b-c)(c-a)
=-(a-b)(c-a)(c-b)/(a-b)(b-c)(c-a)
=1.
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
数学人气:715 ℃时间:2020-03-27 21:15:43
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