cho em như này đúng ko ạ

$A_1=\sqrt{{(3+2\sqrt{2})}^2}=\sqrt{{(3-2\sqrt{2})}^2}$

    $=\left|3+2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|$

$=3+2\sqrt{2}+3-2\sqrt{2}=6$

$A_2=\sqrt{{(5-2\sqrt{6})}^2}-\sqrt{{(5+2\sqrt{6})}^2}$ $=\left|5-2\sqrt{6}\right|-\left|5+2\sqrt{6}\right|$

      $=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}$

$A_3=\sqrt{{(2-\sqrt{3})}^2}+\sqrt{\left(1-\sqrt{3}\right)}$$=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|$

     $=2-\sqrt{3}+\sqrt{3}-1=1$

$A_4=\sqrt{{(3+\sqrt{2})}^2}-\sqrt{\left(1-\sqrt{2}\right)}$$=\left|3+\sqrt{2}\right|-\left|1-\sqrt{2}\right|$ 

     $=3+\sqrt{2}-\left(\sqrt{2}+1\right)=4$2: 

$B_1=(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}):\sqrt{3}$

     $=(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}):\sqrt{3}=-5\sqrt{3}:\sqrt{3}=-5$

$B_2=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}$

     $=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}$

$B_3=2\sqrt{3}-4\sqrt{27}+5\sqrt{48}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=10\sqrt{3}$

$B_4=(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}).\sqrt{2}$

     $=(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}).\sqrt{2}=6\sqrt{2}.\sqrt{2}=12$

cách b