已知实数x,y,z满足x²-6xy+10y²+4y+|z²-3z+2|+4=0,求(x+y

x²-6xy+10y²+4y+|z²-3z+2|+4=0
(x²-6xy+9y²)+(y²+4y+4)+|z²-3z+2|=0
(x-3y)²+(y+2)²)+|z²-3z+2|=0
∴x-3y=0
y+2=0
z²-3z+2=0
解得y=-2,x=-6,z=1或2
(x+y)z=(-2-6)×1=-8
或(x+y)z=(-2-6)×2=-16