根号x+根号y-1+根号z-2=1/2(x+y+z),求x,y,z的值

√x+√(y-1)+√(z-2)=1/2(x+y+z)变形后得[x-2√x+1]+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-2)+1=0即（√x-1）^2+[√(y-1)+1]^2+[√(z-2)-1]^2=0解得x=1y=2z=3