Question

初二分式化简求值

(A-B)/(A+B)+(B-C)/(B+C)+(C-A)/(C+A)+(A-B)(B-C)(C-A)/(A+B)(B+C)(C+A) 拜托啦，急急急！还会追加分的！！

Answer 1

(a-b)/(a+b)+(b-c)/(b+c)+(c-a)/(c+a)+(a-b)(b-c)(c-a)/[(a+b)(b+c)(c+a)]=?

令：x=a+b，y=b+c，z=a+c
--->a-b=z-y，b-c=x-z，c-a=y-x
原式= (z-y)/x+(x-z)/y + (y-x)/z+(z-y)(x-z)(y-x)/(xyz)
= (yz-y²+x²-zx)/(xy) + (y-x)[xy+(z-y)(x-z)]/(xyz)
= (x-y)(x+y-z)/(xy) - (x-y)[xy+(zx-z²-xy+yz)]/(xyz)
= (x-y)(zx+yz-z²)/(xyz) - (x-y)(zx-z²+yz)/(xyz)
= 0