Question

计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y) 求速度

Answer 1

(x-y)^2/(z-x)(z-y)+(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z)=[-(x-y)^3-(y-z)^3-(z-x)^3]/[(x-y)(y-z)(z-x)]=[-x^3+3x^2y-3xy^2+y^3-y^3+3y^2z-3yz^2+z^3-z^3+3z^2x-3zx^2+x^3]/[(x-y)(y-z)(z-x)]=[3x^2y-3xy^2+3y^2z-3yz^2+3z^2x-3zx^2]/[(x-y)(y-z)(z-x)]=3[x^2y-xy^2+y^2z-yz^2+z^2x-zx^2]/[(x-y)(y-z)(z-x)]=3[xy(x-y)+yz(y-z)+zx(z-x)]/[(x-y)(y-z)(z-x)]只能这样了，去掉立方项

追问

不好意思，我要的不是这道题，麻烦你看清题目

Answer 2

答：

x^2/[(x-y)(x-z)]+y^2/[(y-x)(y-z)]+z^2/[(z-x)(z-y)]
=[(y-z)x^2-(x-z)y^2+(x-y)z^2] / [(x-y)(x-z)(y-z)]
=[(y-z)x^2-(y-z+x-y)y^2+(x-y)z^2] / [(x-y)(x-z)(y-z)]
=[(y-z)(x^2-y^2)+(x-y)(z^2-y^)] / [(x-y)(x-z)(y-z)]
=[(y-z)(x-y)(x+y)+(x-y)(z-y)(z+y)] / [(x-y)(x-z)(y-z)]
=[(x+y)-(y+z)] / (x-z)
=(x-z)/(x-z)
=1

追问

第三步是什么意思

追答

第三步是关键：x-z=(y-z)+(x-y)，为后面的分解利用y-z和x-y

追问

哦，那第4步怎么变成第5步

追答

把公因式约掉~~~