作业帮分式化简(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]}，
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/12 15:37:22
![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]}，](/uploads/image/z/2576782-46-2.jpg?t=%E5%88%86%E5%BC%8F%E5%8C%96%E7%AE%80%28x%5E2%2Byz%29%2F%28x%5E2%2B%28y-x%29x-yz%29%2B%28y%5E2-zx%29%2F%28y%5E2%2B%28z%2Bx%29y%2Bzx%29%2B%28z%5E2%3Dxy%29%2F%28z%5E2-%28x-y%29z-xy%29%E5%86%8D%E4%BB%8E%E8%BE%93%E4%B8%80%E9%81%8D%E5%A5%BD%E4%BA%86%EF%BC%8C%7B%28x%5E2%2Byz%29%C3%B7%5Bx%5E2%2B%28y-z%29x-yz%5D%7D%2B%7B%28y%5E2-zx%29%C3%B7%5By%5E2%2B%28z%2Bx%29y-zx%5D%7D%2B%7B%28z%5E2%2Bxy%29%C3%B7%5Bz%5E2-%28x-y%29z-xy%5D%7D%EF%BC%8C)
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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