3/ $A\,=\sqrt{4-2\sqrt 3}-\sqrt 3\\\quad =\sqrt{3-2\sqrt 3+1}-\sqrt 3\\\quad =\sqrt{(\sqrt 3)^2-2.\sqrt 3.1+1^2}-\sqrt 3\\\quad =\sqrt{(\sqrt 3-1)^2}-\sqrt 3\\\quad =|\sqrt 3-1|-\sqrt 3\\\quad =\sqrt 3-1-\sqrt 3(vì\,\,\sqrt 3-1>0)\\\quad =-1$
Vậy $A=-1$
$B\,=\sqrt{6+2\sqrt 5}-\sqrt 5\\\quad =\sqrt{5+2\sqrt 5+1}-\sqrt 5\\\quad =\sqrt{(\sqrt 5)^2+2.\sqrt 5.1+1^2}-\sqrt 5\\\quad =\sqrt{(\sqrt 5+1)^2}-\sqrt 5\\\quad =|\sqrt 5+1|-\sqrt 5\\\quad =\sqrt 5+1-\sqrt 5(vì\,\,\sqrt 5+1>0)\\\quad =1$
Vậy $B=1$
$C\,=\sqrt{17-12\sqrt 2}+\sqrt 8\\\quad =\sqrt{17-2.6.\sqrt 2}+\sqrt 8\\\quad =\sqrt{17-2.3.2\sqrt 2}+\sqrt 8\\\quad =\sqrt{9-2.3.2\sqrt 2+8}+2\sqrt 2\\\quad =\sqrt{3^2-2.3.2\sqrt 2+(2\sqrt 2)^2}+2\sqrt 2\\\quad =\sqrt{(3-2\sqrt 2)^2}+2\sqrt 2\\\quad =|3-2\sqrt 2|+2\sqrt 2\\\quad =3-2\sqrt 2+2\sqrt 2(vì\,\,3-2\sqrt 2>0)\\\quad =3$
Vậy $C=3$
4/ $A\,=\sqrt{6+2\sqrt{4-2\sqrt 3}}\\\quad =\sqrt{6+2\sqrt{3-2\sqrt 3+1}}\\\quad =\sqrt{6+2\sqrt{(\sqrt 3)^2-2.\sqrt 3.1+1^2}}\\\quad =\sqrt{6+2\sqrt{(\sqrt 3-1)^2}}\\\quad =\sqrt{6+2|\sqrt 3-1|}\\\quad =\sqrt{6+2(\sqrt 3-1)}(vì\,\,\sqrt 3-1>0)\\\quad =\sqrt{6+2\sqrt 3-2}\\\quad =\sqrt{4+2\sqrt 3}\\\quad =\sqrt{3+2\sqrt 3+1}\\\quad =\sqrt{(\sqrt 3)^2+2.\sqrt 3.1+1^2}\\\quad =\sqrt{(\sqrt 3+1)^2}\\\quad =|\sqrt 3+1|\\\quad =\sqrt 3+1(vì\,\,\sqrt 3+1>0)$
Vậy $A=\sqrt 3+1$
$B\,=\sqrt{6-2\sqrt{4+2\sqrt 3}}\\\quad =\sqrt{6-2\sqrt{3+2\sqrt 3+1}}\\\quad =\sqrt{6-2\sqrt{(\sqrt 3)^2+2.\sqrt 3.1+1^2}}\\\quad =\sqrt{6-2\sqrt{(\sqrt 3+1)^2}}\\\quad =\sqrt{6-2|\sqrt 3+1|}\\\quad =\sqrt{6-2(\sqrt 3+1)}(vì\,\,\sqrt 3+1>0)\\\quad =\sqrt{6-2\sqrt 3-2}\\\quad =\sqrt{4-2\sqrt 3}\\\quad=\sqrt{3-2\sqrt 3+1}\\\quad =\sqrt{(\sqrt 3)^2-2.\sqrt 3.1+1^2}\\\quad =\sqrt{(\sqrt 3-1)^2}\\\quad =|\sqrt 3-1|\\\quad =\sqrt 3-1(vì\,\,\sqrt 3-1>0)$
Vậy $B=\sqrt 3-1$
$C\,=\sqrt{7+4\sqrt{21-12\sqrt 3}}\\\quad =\sqrt{7+4\sqrt{21-2.3.2\sqrt 3}}\\\quad =\sqrt{7+4\sqrt{12-2.2\sqrt 3.3+9}}\\\quad =\sqrt{7+4\sqrt{(2\sqrt3)^2-2.2\sqrt 3.3+3^2}}\\\quad =\sqrt{7+4\sqrt{(2\sqrt 3-3)^2}}\\\quad =\sqrt{7+4|2\sqrt 3-3|}\\\quad =\sqrt{7+4(2\sqrt 3-3)}(vì\,\,2\sqrt 3-3>0)\\\quad=\sqrt{7+8\sqrt 3-12}\\\quad =\sqrt{-5+8\sqrt 3}\\\quad =\sqrt{8\sqrt 3-3}$
Vậy $C=\sqrt{8\sqrt 3-5}$
