Question

一道初中代数题，高手帮帮忙啊

证明：（a-b/a+b）+（b-c/b+c）+（c-a/c+a）+(a-b)(b-c)(c-a)/(a+b)(b+c)(c+a)=0
过程要详细，好的加分！！！
有没有人来帮帮我啊，急啊！！！

Answer 1

解：设：x=a+b，y=b+c，m=c+a
则a-b=m-y，b-c=x-m，c-a=y-x
(a-b)/(a+b)=(m-y)/x (b-c)/(b+c)=(x-m)/y (c-a)/(c+a)=(y-x)/m
得：原式= (m-y)/x +(x-m)/y+ (y-x)/m+(m-y)(x-m)(y-x)/(xym)
　　= (my-y²+x²-mx)/(xy) + (y-x)[xy+(m-y)(x-m)]/(xym)
　　= (x-y)(x+y-m)/(xy) - (x-y)[xy+(mx-m²-xy+my)]/(xym)
= (x-y)(mx+ym-m²)/(xym) - (x-y)(mx-m²+ym)/(xym)
　　= 0
如果你不想用这种方法，你就要通分消去分母计算，等式两边同时乘以(a+b)(b+c)(c+a)，很麻烦。

追问

(m-y)/x +(x-m)/y+ (y-x)/m+(m-y)(x-m)(y-x)/(xym)
　　= (my-y²+x²-mx)/(xy) + (y-x)[xy+(m-y)(x-m)]/(xym)
这一步没看明白。。。

追答

(m-y)/x +(x-m)/y+ (y-x)/m+(m-y)(x-m)(y-x)/(xym)
=(m-y)/x ×y/y+(x-m)/y×x/x+ (y-x)/m×xy/xy+(m-y)(x-m)(y-x)/(xym)
=(m-y)y/xy +(x-m)x/xy+ xy(y-x)/xym+(m-y)(x-m)(y-x)/(xym)
=(my-y²)/xy +(x²-mx)/xy+ xy(y-x)/xym+(m-y)(x-m)(y-x)/(xym)
把(my-y²)/xy +(x²-mx)/xy合并得(my-y²+x²-mx)/(xy)
把xy(y-x)/xym和(m-y)(x-m)(y-x)/(xym) 中的(y-x)提出来得(y-x)[xy/xym+(m-y)(x-m)/(xym)]，之后再把xy/xym+(m-y)(x-m)/(xym)合并得： (y-x)[xy+(m-y)(x-m)]/(xym)
最后把(my-y²+x²-mx)/(xy)和 (y-x)[xy+(m-y)(x-m)]/(xym) 相加
得(my-y²+x²-mx)/(xy) + (y-x)[xy+(m-y)(x-m)]/(xym)

Answer 2

前三项通分，分子化简
后一项把分子拆开，就OK了

追问

能在详细点么

Answer 3

容我想想先