若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx一、若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx.（2）.2分之1[(x-y)²+(y-z)²(z-x)²].二、已知（x+2）²+|x+y+5|=0,求3x&

若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx一、若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx.（2）.2分之1[(x-y)²+(y-z)²(z-x)²].二、已知（x+2）²+|x+y+5|=0,求3x&
若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx
一、若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx.
（2）.2分之1[(x-y)²+(y-z)²(z-x)²].
二、已知（x+2）²+|x+y+5|=0,求3x²y+{-2x²y-[-2xy+(x²y-4x²)]-xy}
已经很晚、我明天还要交作业,

若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx一、若(x-3)²+|y+1|+z²=0.求：(1).x²+y²+z²-xy-yz-zx.（2）.2分之1[(x-y)²+(y-z)²(z-x)²].二、已知（x+2）²+|x+y+5|=0,求3x&
一、因为 (x-3)²+|y+1|+z²=0
所以 (x-3)²=0
|y+1|=0
z²=0
所以：x=3
y=-1
z=0
(1).x²+y²+z²-xy-yz-zx
=9+1+0-(-3)-0-0
=13
(2)[(x-y)²+(y-z)²(z-x)²]/2
=4²+1²+1²
=18
二、因为 （x+2）²+|x+y+5|=0
所以 （x+2）²=0
|x+y+5|=0
所以 x=-2
x+y=-5
y=-3
将x、y代入,自己计算
3x²y+{-2x²y-[-2xy+(x²y-4x²)]-xy}
=.