已知x,y属于(0,正无穷),且满足xy=x+y+3,求xy的最小值.

设t=xy
则：x=t/y
xy=x+y+3
t=t/y+y+3
y^2+(3-t)y+t=0
△=(3-t)^2-4t
=9-10t+t^2
=(t-1)(t-9)≥0
t≥9,或,t≤1
因为x,y大于0,所以,y1+y2=t-3>0
t>3
所以,t≥9
即：xy的最小值为9