问题
填空题
设x、y、z满足关系式x-1=
|
答案
令x-1=
y+1 |
2 |
z-2 |
3 |
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 |
设x、y、z满足关系式x-1=
|
令x-1=
y+1 |
2 |
z-2 |
3 |
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 |