已知x满足不等式：2（log1/2 x）²+7log1/2 x +3≤0,则函数f(x)=(log2 x/4)·(log2 x/2)的最大值和最小值分别为

令t=log1/2 x
所以
2（log1/2 x）²+7log1/2 x +3≤0
变为
2t^2+7t+3最小不应该是3/4么。为啥？代入x=2^(3/2)=(1/2)^(-3/2)(log2 x/4)·(log2 x/2)=(log2 x-log 2 4)(log 2 x- log 2 2)=(3/2-2)(3/2-1)=(-1/2)(1/2)=-1/42（log1/2 x）²+7log1/2 x +3=2(-3/2)^2+7(-3/2)+3=9/2-21/2+3=-3<0成立所以-1/4没问题