A=đề
=x^4+4x^3+4x^2+2x^2+4x+1+3
=(x^2+2x)^2+2(x^2+2x)+1+3
=(x^2+2x+1)^2+3
(x^2+2x+1)^2>=0 =>A>=3
Dấu '=' xảy ra<=>x^2+2x+1=0
<=>(x+1)^2=0
<=>x+1=0
<=>x=-1
B=đề
=x^4-4x^3+4x^2-2x^2+4x+1+4
=(x^2-2x)^2-2(x^2-2x)+1+4
=(x^2-2x-1)^2+4
(x^2-2x-1)^2>=0=>B>=4
Dấu '=' xảy ra<=>x^2-2x-1=0
<=>x^2-2x+1=2
<=>(x-1)^2=2
<=>x=±√2+1