第二个分母写错了?
(y - x)(z - x)/(x - 2y + z)/(x + y - 2z) + (z - y)(x - y)/(x + y - 2z)/(y + z - 2x)
+ (x - z)(y - z)/(y + z - 2x)/(x - 2y + z) =1
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
速速回答
速速回答
数学人气：241 ℃时间：2020-01-30 08:15:37
优质解答
我来回答
类似推荐
- 化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+（z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
- 计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
- 化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)
- 已知xyz=1，x+y+z=2，x2+y2+z2=16，求代数式1/xy+2z+1/yz+2x+1/zx+2y的值．
- 设x,y,z是实数,且(x-y)^2+(y-z)^2+(z-x)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2求[(xy+1)(yz+1)(zx+1)]/