میدانیم $\sqrt {\frac{a}{b}} = \frac{{\sqrt a }}{{\sqrt b }}$ و $\sqrt {{x^2}} = \,\left| x \right.)$
$\sqrt 8 = \sqrt {4 \times 2} = \sqrt 4 \times \sqrt 2 = 2 \times \sqrt 2 = 2\sqrt 2 $
$\frac{{\sqrt {\frac{3}{2}} + \frac{1}{{\sqrt 2 }}}}{{\sqrt {{{(1 - \sqrt 2 )}^2}} }} \times \frac{{\sqrt 8 }}{{1 + \sqrt 2 }} = \frac{{\frac{{\sqrt 3 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}}}{{\left| {1 - \sqrt 2 } \right|}} \times \frac{{2\sqrt 2 }}{{1 + \sqrt 2 }} = \frac{{\frac{{\sqrt 3 + 1}}{{\sqrt 2 }}}}{{ - (1 - \sqrt 2 )}} \times \frac{{2\sqrt 2 }}{{1 + \sqrt 2 }}$
$ = \frac{{\sqrt 3 + 1}}{{ - \sqrt 2 (1 - \sqrt 2 )}} \times \frac{{2\sqrt 2 }}{{1 + \sqrt 2 }} = \frac{{2(\sqrt 3 + 1)}}{{ - (1 - 3)}} = + 2(\sqrt 3 + 1)$