题目内容
若x,y,z均为非负数,且满足x-1=
=
,则x2+y2+z2可取得的最小值为( )
(提示:令x-1=
=
=t)
y+1 |
2 |
z-2 |
3 |
(提示:令x-1=
y+1 |
2 |
z-2 |
3 |
A.3 | B.
| C.0 | D.
|
令x-1=
=
=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,
≥
,
≥-
,
∵x-1=
=
=t,
∴y≥
,
∴当t=
时,其最小值=14×
+10×
+6=
.
故选D.
y+1 |
2 |
z-2 |
3 |
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 |
∴y≥
1 |
2 |
∴当t=
1 |
2 |
1 |
4 |
1 |
2 |
29 |
2 |
故选D.
练习册系列答案
相关题目