(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x+z-2y)的值我已经做到这了：设(X-Y)为a （Z-Y)为b (Z-X)为c则原题= (-a-c)/-(a+b)(b+c)+ab/-(c+b)(a+c)+c-b/(-a+c)(a+b)

来源：学生作业帮助网编辑：作业帮时间：2024/11/14 14:37:48

(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x+z-2y)的值我已经做到这了：设(X-Y)为a （Z-Y)为b (Z-X)为c则原题= (-a*-c)/-(a+b)(b+c)+ab/-(c+b)(a+c)+c*-b/(-a+c)(a+b)
(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x+z-2y)的值
我已经做到这了：设(X-Y)为a （Z-Y)为b (Z-X)为c
则原题= (-a*-c)/-(a+b)(b+c)+ab/-(c+b)(a+c)+c*-b/(-a+c)(a+b)

(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x+z-2y)的值我已经做到这了：设(X-Y)为a （Z-Y)为b (Z-X)为c则原题= (-a*-c)/-(a+b)(b+c)+ab/-(c+b)(a+c)+c*-b/(-a+c)(a+b)
∑是循环和
例如
∑a=a+b+c
∑a^2=a^2+b^2+c^2
∑(z-y)(x-y)/(x+y-2z)(y+z-2x)
＝∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z-2y)
＝[1/(x+y-2z)(y+z-2x)(x+z-2y)]∑(z-y)(x-y)(x+z-2y)
你的设法不科学,丧失了轮转对称性!
x-y=a
y-z=b
z-x=c
a+b+c=0
∑(z-y)(x-y)(x+z-2y)
=∑(z-y)(x-y)(x-y)+∑(z-y)(x-y)(z-y)
=-∑aab+∑bba
=(a-b)(b-c)(c-a)
1/(x+y-2z)(y+z-2x)(x+z-2y)
=1/(b-c)(c-a)(a-b)
所求＝1

接刚才的，=ac(c-a）+ab(b-c)+bc(c-a)/(a-b)（b-c)(c-a）

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