分式化简(x^2+yz)/(x^2+(y-x)x-yz)+(y^2-zx)/(y^2+(z+x)y+zx)+(z^2=xy)/(z^2-(x-y)z-xy)再从输一遍好了，{(x^2+yz)÷[x^2+(y-z)x-yz]}+{(y^2-zx)÷[y^2+(z+x)y-zx]}+{(z^2+xy)÷[z^2-(x-y)z-xy]}，

分式化简(x^2+yz)/(x^2+(y-x)x-yz)+(y^2-zx)/(y^2+(z+x)y+zx)+(z^2=xy)/(z^2-(x-y)z-xy)再从输一遍好了，{(x^2+yz)÷[x^2+(y-z)x-yz]}+{(y^2-zx)÷[y^2+(z+x)y-zx]}+{(z^2+xy)÷[z^2-(x-y)z-xy]}，
分式化简(x^2+yz)/(x^2+(y-x)x-yz)+(y^2-zx)/(y^2+(z+x)y+zx)+(z^2=xy)/(z^2-(x-y)z-xy)
再从输一遍好了，{(x^2+yz)÷[x^2+(y-z)x-yz]}+{(y^2-zx)÷[y^2+(z+x)y-zx]}+{(z^2+xy)÷[z^2-(x-y)z-xy]}，

分式化简(x^2+yz)/(x^2+(y-x)x-yz)+(y^2-zx)/(y^2+(z+x)y+zx)+(z^2=xy)/(z^2-(x-y)z-xy)再从输一遍好了，{(x^2+yz)÷[x^2+(y-z)x-yz]}+{(y^2-zx)÷[y^2+(z+x)y-zx]}+{(z^2+xy)÷[z^2-(x-y)z-xy]}，
题目确定是这样吗?如果y^2+(z+x)y-zx改为y^2+(z+x)y+zx
那么
三个分母分别因式分解为：（x+y)(x-y),(y+x)(y+z),(y+z)(z-x)
然后再通分,分子展开后正好为0
所以原式＝0