Ta có:
$\sqrt{2 + \sqrt3}.\sqrt{2 + \sqrt{2 +\sqrt3}}.\sqrt{2 + \sqrt{2 + \sqrt{2 +\sqrt3}}}.\sqrt{2 -\sqrt{2 + \sqrt{2 +\sqrt3}}}$
$= \sqrt{2 + \sqrt3}.\sqrt{2 + \sqrt{2 +\sqrt3}}.\sqrt{\left(2 + \sqrt{2 + \sqrt{2 +\sqrt3}}\right)\left(2 + \sqrt{2 + \sqrt{2 +\sqrt3}}\right)}$
$= \sqrt{2 + \sqrt3}.\sqrt{2 + \sqrt{2 +\sqrt3}}.\sqrt{4 - \left(2 + \sqrt{2 +\sqrt3}\right)}$
$= \sqrt{2 + \sqrt3}.\sqrt{2 + \sqrt{2 +\sqrt3}}.\sqrt{2- \sqrt{2 +\sqrt3}}$
$= \sqrt{2 + \sqrt3}.\sqrt{\left(2 + \sqrt{2 +\sqrt3}\right)\left(2 -\sqrt{2 +\sqrt3}\right)}$
$= \sqrt{2 + \sqrt3}.\sqrt{4 - \left(2 +\sqrt3\right)}$
$= \sqrt{2 +\sqrt3}.\sqrt{2 -\sqrt3}$
$=\sqrt{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}$
$=\sqrt{4 -3}$
$= 1$
