问题 填空题
设x、y、z满足关系式x-1=
y+1
2
=
z-2
3
,则x2+y2+z2的最小值为______.
答案

令x-1=

y+1
2
=
z-2
3
=k,则x=k+1,y=2k-1,z=3k+2,

于是x2+y2+z2=(k+1)2+(2k-1)2+(3k+2)2

=k2+2k+1+4k2+1-4k+9k2+4+12k

=14k2+10k+6,

其最小值为

4ac-b2
4a
=
4×14×6-100
4×14
=
59
14

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