$A=\dfrac{\sqrt{ \sqrt5+2}+\sqrt{ \sqrt5-2} }{\sqrt{ \sqrt5+1}}$
Xét $A_1=\sqrt{ \sqrt5+2}+\sqrt{ \sqrt5-2}$
$A_1^2=\sqrt5+2+\sqrt5-2+2\sqrt{(\sqrt5+2)(\sqrt5-2)}$
$=2\sqrt5+2\sqrt{5-4}$
$=2(\sqrt5+1)$
$\to A_1=\sqrt2.\sqrt{\sqrt5+1}$
$\to A=\dfrac{A_1}{\sqrt{ \sqrt5+1}}=\sqrt2$
Vậy $A=\sqrt2$
$B=\sqrt{3-2\sqrt2}=\sqrt{(\sqrt2-1)^2}=|\sqrt2-1|=\sqrt2-1$
(bạn xem lại đề vì $A\ne B$)