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

来源：学生作业帮编辑：百度作业网作业帮分类：数学作业时间：2024/07/11 20:30:39

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
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(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x (x-2y+z)(x+y-2z)分之（y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x- 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) (x+y+z)^2-(x-y-z)^2 分式约分:(y+z-x)/{x^2-(y+z)^2} (x+y-z)^2-(x-y+z)^2=? (x+y-z)(x-y+z)= 设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+( x,y,z为实数且(y-z)^2+(x-y)^2+(z-x)^2=(y+z-2x)^2+(x+z-2y)^2+(x+y