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

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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的值已知x+y+z=6,1/(x-1)+1/(y-2)+1/(z-3)=0,求(

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
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

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
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.

x+y+z=6，得到(x－1)＋(y－2)＋(z－3)=0，且1/(x-1)+1/(y-2)+1/(z-3)=0，等介于a＋b＋c=0且1/a＋1/b＋1/c=0，由第二个去分母得到：ab＋bc＋ca=0，第一个平方后再减去第二个的2倍，得：a²＋b²＋c²=0。
题目好像有问题啊。。

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