若x≤0,则y=f[f(x)]-1=2^(2^x)-1令2^(2^x)-1=0即2^(2^x)=2^0即2^x=0,显然无解表明此时y=f[f(x)]-1无零点若x>0,则y=f[f(x)]-1=log2[log2(x)]-1令log2[log2(x)]-1=0即log2[log2(x)]=log2(2)即log2(x)=2解得x=4所以y=f[...