已知X、Y都在区间（-2,2）内,且XY=1,则U=4/（4-X^2）+9/(9-Y^2)的最小值是多少?

由于:X,Y属于(-2,2)
则有：
4-x^2>0,9-y^2>0
又：XY=-1
由均值不等式,得：
u=4/（4-X^2）+9/(9-Y^2)
≥2√[4/(4-x^2)*9/(9-y^2)]
=12/√[(36-9x^2-4y^2+(xy)^2]
=12/√[37-(9x^2+4y^2)]
≥12/√[37-2√(36x^2y^2)]
=12/√[37-2*6*1]
=12/5
当且仅当4/(4-X^2)=9/(9-Y^2),
即：x=√(2/3),y=-√(3/2)时,
u取最小值=12/5