f(x)=(log2x-log22)(log2x-log24)=(log2x-1)( log2x-2)设t=log2x,根号2≤x≤4,则1/2≤t≤2.得f(x)=(t-1)( t -2)=t^2-3t+2=(t-3/2)^2-1/4所以ymin=y(3/2)=-1/4,此时t=3/2,则log2x=3/2,x=2^(3/2)=2√2.ymax=y(1/2)=3/4...