y=a^(x^2-3x+3) (a>0且不=1),当1《x《3时，y最小=8，求a

y=a^(x^2-3x+3)=a^[(x-3/2)^2+3/4],
若0<a<1,则有y(3)=a^3=8 a=2 矛盾
若a>1,则有y(3/2)=a^(3/4)=8 a=16

故a=16

y=a^(x^2-3x+3) (a>0且≠1),当1<=x<=3时，y最小=8
y=a((x-3/2)^2+3-9/4)
=a(x-3/2)^2+3a/4
ymin最小=8
x=3/2
3a/4=8
a=32/3