a/ $A=x^2-4x+1\\=x^2-4x+4-3\\=(x-2)^2-3$
Vì $(x-2)^2\ge 0→A≥-3\\→\min A=-3$
$→$ Dấu "=" xảy ra khi $x-2=0↔x=2$
Vậy $\min A=-3$ khi $x=2$
b/ $B=4x-x^2+1\\=-x^2+4x-4+5\\-(x-2)^2+5$
Vì $-(x-2)^2\le 0→B≤5\\→\max B=5$
$→$ Dấu "=" xảy ra khi $x-2=0↔x=2$
Vậy $\max B=5$ khi $x=2$