已知x≥5/2,则f(x)=[x^2-4x+5]/[2x-4]等于多少,

问题描述:

已知x≥5/2,则f(x)=[x^2-4x+5]/[2x-4]等于多少,

f(x)=[x^2-4x+5]/[2x-4]
=[(x-2)^2+1]/2(x-2)
=(x-2)/2+1/2(x-2)≥1
(x-2)=1/(x-2)
(x-2)=1
x=3 因为x≥5/2
最小值能达到
所以 f(x)=[x^2-4x+5]/[2x-4]≥1