Đáp án: $=-\sqrt{6}$
Giải thích các bước giải:
Đặt $A=\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}$
$⇒A\sqrt{2}=\sqrt{2}.\sqrt{4-\sqrt{15}}-\sqrt{2}.\sqrt{4+\sqrt{15}}$
$=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}$
$=\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}-\sqrt{5+2.\sqrt{5}.\sqrt{3}+3}$
$=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}$
$=|\sqrt{5}-\sqrt{3}|-|\sqrt{5}+\sqrt{3}|$
$=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})$
$=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}$
$⇒A=-\sqrt{2}.\sqrt{3}=-\sqrt{6}$
