Đặt $A = \sqrt{6+2\sqrt{3}+2\sqrt{5}+\sqrt{15}}-\sqrt{4+\sqrt{15}}$
$\to A\sqrt2 = \sqrt{12+4\sqrt{3}+4\sqrt{5}+2\sqrt{15}}-\sqrt{8+2\sqrt{15}}$
$\to A\sqrt2= \sqrt{\left(\sqrt5\right)^2 + \left(\sqrt3\right)^2 + 2^2 +2.2.\sqrt{3}+2.2.\sqrt{5}+2\sqrt{3}.\sqrt5}-\sqrt{\left(\sqrt5\right)^2+2\sqrt{5}.\sqrt3 + \left(\sqrt3\right)^2}$
$\to A\sqrt2 = \sqrt{\left(2 +\sqrt3 +\sqrt5\right)^2} - \sqrt{\left(\sqrt3 +\sqrt5\right)^2}$
$\to A\sqrt2 = 2 + \sqrt3 +\sqrt5 - \left(\sqrt3 +\sqrt5\right)$
$\to A\sqrt2 = 2$
$\to A =\sqrt2$
Vậy $\sqrt{6+2\sqrt{3}+2\sqrt{5}+\sqrt{15}}-\sqrt{4+\sqrt{15}}=\sqrt2$
