分式化简(x^2+yz)/(x^2+(y-x)x-yz)+(y^2-zx)/(y^2+(z+x)y+zx)+(z^2=x
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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]},
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=x](/uploads/image/z/17208396-36-6.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%3Dx)
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
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
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