令log2(x)=t2[log0.5(x)]²-14log4(x)+3≤0由對(duì)數(shù)換底公式 得：2[-log2(x)]²-14[log(x/log2(4)]+3≤02t²-7t+3≤0==>1≤t≤3/2f(x)=log2(x/2) log√2(√x/2)=[log2(x)-1]*{[log2(√x/2)]/log2(√2)]}y=...答案最大值是2，最小值是-1/4不好意思，第七行錯(cuò)了，更正如下 ： 令log2(x)=t2[log0.5(x)]²-14log4(x)+3≤0由對(duì)數(shù)換底公式 得：2[-log2(x)]²-14[log(x/log2(4)]+3≤02t²-7t+3≤0(2t-1)(t-3)≤01/2≤t≤3f(x)=log2(x/2)* log√2(√x/2)=log2(x/2)=log2(x)-1=(t-1) log√2(√x/2)=[log2(√x/2)]/[log2(√2)]=2log2(√x/2)=2[log2(√x)-1]=2(1/2t-1)=(t-2)所以,f(x)=log2(x/2)* log√2(√x/2)=(t-1)(t-2) =t²-3t+2函數(shù) t²-3t+2的對(duì)稱軸為t=3/2,在【1/2,3]上先減后增，所以函數(shù)當(dāng)t=3/2時(shí)最小，最小值為 9/4-9/2+2=-1/4當(dāng)t=3時(shí)，最大最大值為：3²-3*3+2=2