作业帮若x-1=(y+1)/2=(z-2)/3,求x^2-y^2+z^2的最小值.
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若x-1=(y+1)/2=(z-2)/3,求x^2-y^2+z^2的最小值.
若x-1=(y+1)/2=(z-2)/3,求x^2-y^2+z^2的最小值.
若x-1=(y+1)/2=(z-2)/3,求x^2-y^2+z^2的最小值.
设x-1=(y+1)/2=(z-2)/3=t,
x=t+1,y=2t-1,z=3t+2
x^2-y^2+z^2
=(t+1)^2-(2t-1)^2+(3t+2)^2
=6t^2+18t+4
=6(t+3/2)^2-19/2
所以最小值是-19/2
x-1=(y+1)/2=(z-2)/3 =k
x= k+1, y= 2k-1, z=3k+2
S = x^2-y^2+z^2
= (k+1)^2-(2k-1)^2+(3k+2)^2
=6k^2+ 10k +4
S' =12k+10 =0
k=-5/6
S''=12>0 (min)
min S =6(-5/6)^2+ 10(-5/6) +4 = 25/6- 50/6 +4 = -1/6
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