（a-b/a+b)+(b-c/b+c)+(c-a/c+a)+[(a-b)(b-c)(c-a)/(a+
（a-b/a+b)+(b-c/b+c)+(c-a/c+a)+[(a-b)(b-c)(c-a)/(a+b)(b+c)(c +a)}=?
(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
答：化简(b－c)^2/4＝(a－b)(c－a)，得 [(a－b)+(c－a)]^2/4＝(a－b)(c－a) [(a－b)+(c－a)]^2＝4(a－b)(c－a...详情>>
答：详情>>
