若f（x）=x^2-x+b,且f（log2（a））=b,log2[f(a)]=2(a不等于1),求f（log2（x））的最小值及对应的X值,

f（x）=x^2-x+bf(log2 a)=log²2 a-log2 a+b=blog2 a*(log2 a-1)=0所以log2 a=0 得 a=1(舍去)或log2 a=1 得a=2log2 f(a)=log2 f(2)=log2 (4-2+b)=22+b=4所以b=2则f(log2 x)=log²2 x-log2 x+2=[log2 x-1/2]&...=[log2 x-1/2]²+7/4怎么得到这一步的？配方呢则f(log2 x)=log²2 x-log2 x+2=[log2 x-1/2]²+2-1/4=[log2 x-1/2]²+7/4