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