已知x+y+z=6,1/(x-1)+1/(y-2)+1/(z-3)=0,求(x-1)^2+(y-2)^2+(z-3)^2的值
问题描述：
已知x+y+z=6,1/(x-1)+1/(y-2)+1/(z-3)=0,求(x-1)^2+(y-2)^2+(z-3)^2的值
答
为了方便输入,设x-1=a,y-2=b,z-3=c
由x+y+z=6,得a+b+c=0
由1/a+1/b+1/c=0得bc+ac+ab=0
(a+b+c)^2=a^2+b^2+c^2+2(bc+ac+ab)
可得所求值为(x-1)^2+(y-2)^2+(z-3)^2=a^2+b^2+c^2=0.