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若x,y,z均为非负数,且满足x-1=y+12=z-23,则x2+y2+z2可取得的最小值为()(提示:令x-1=y+12=z-23=t)A.3B.5914C.0D.292

题目详情
若x,y,z均为非负数,且满足x-1=
y+1
2
=
z-2
3
,则x2+y2+z2可取得的最小值为(  )
(提示:令x-1=
y+1
2
=
z-2
3
=t)
A. 3
B.
59
14

C. 0
D.
29
2
▼优质解答
答案和解析
x-1=
y+1
2
=
z-2
3
=t,
则x=t+1,y=2t-1,z=3t+2,
于是x2+y2+z2=(t+1)2+(2t-1)2+(3t+2)2
=t2+2t+1+4t2+1-4t+9t2+4+12t
=14t2+10t+6,
∵x,y,z均为非负数,
∴x-1≥-1,
y+1
2
1
2
z-2
3
≥-
2
3

x-1=
y+1
2
=
z-2
3
=t,
∴t≥
1
2

∴当t=
1
2
时,其最小值=14×
1
4
+10×
1
2
+6=
29
2

故选D.