化简:分式加减化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-y-x)/(
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 11:55:44
化简:分式加减
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-y-x)/(z-y)(z-x)
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-y-x)/(z-y)(z-x)
(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-y-x)/(z-y)(z-x)
= (x-y+x-z)/(x-y)(x-z)+(y-z+y-x)/(y-z)(y-x)+(z-y+z-x)/(z-y)(z-x)
= (x-y)/(x-y)(x-z)+(x-z)/(x-y)(x-z)+
(y-z)/(y-z)(y-x)+(y-x)/(y-z)(y-x)+
(z-y)/(z-y)(z-x)+(z-x)/(z-y)(z-x)
= 1/(x-z)+1/(x-y)+1/(y-x)+1/(y-z)+1/(z-x)+1/(z-y)
= 1/(x-y)+1/(y-x)+1/(x-z)+1/(z-x)+1/(z-y)+1/(y-z)
=0
= (x-y+x-z)/(x-y)(x-z)+(y-z+y-x)/(y-z)(y-x)+(z-y+z-x)/(z-y)(z-x)
= (x-y)/(x-y)(x-z)+(x-z)/(x-y)(x-z)+
(y-z)/(y-z)(y-x)+(y-x)/(y-z)(y-x)+
(z-y)/(z-y)(z-x)+(z-x)/(z-y)(z-x)
= 1/(x-z)+1/(x-y)+1/(y-x)+1/(y-z)+1/(z-x)+1/(z-y)
= 1/(x-y)+1/(y-x)+1/(x-z)+1/(z-x)+1/(z-y)+1/(y-z)
=0
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x
分式约分:(y+z-x)/{x^2-(y+z)^2}
(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)/(xy-2z)(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)
如何化简(x+y+z)(x+y-z)(x-y+z)(-x+y+z)
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)(x-y+z)=
(x+y+z)^2-(x-y-z)^2
(x+y-z)^2-(x-y+z)^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