Đáp án:
$\begin{array}{l}
a)\\
\sqrt {2{x^2} - \sqrt 2 .x + \dfrac{1}{4}} = \sqrt 2 x\\
\Leftrightarrow \sqrt {{{\left( {\sqrt 2 x} \right)}^2} - 2.\sqrt 2 x.\dfrac{1}{2} + \dfrac{1}{4}} = \sqrt 2 x\\
\Leftrightarrow \sqrt {{{\left( {\sqrt 2 x - \dfrac{1}{2}} \right)}^2}} = \sqrt 2 x\\
\Leftrightarrow \left| {\sqrt 2 x - \dfrac{1}{2}} \right| = \sqrt 2 x\\
\Leftrightarrow \sqrt 2 x - \dfrac{1}{2} = - \sqrt 2 x\\
\Leftrightarrow 2\sqrt 2 x = \dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{1}{{4\sqrt 2 }} = \dfrac{{\sqrt 2 }}{8}\\
Vậy\,x = \dfrac{{\sqrt 2 }}{8}\\
b)Dkxd:x \ge - 2\\
\sqrt {4x + 8} + \dfrac{1}{3}\sqrt {9x + 18} = 3\sqrt {\dfrac{{x + 2}}{4}} + \sqrt 2 \\
\Leftrightarrow 2\sqrt {x + 2} + \dfrac{1}{3}.3\sqrt {x + 2} = \dfrac{3}{2}\sqrt {x + 2} + \sqrt 2 \\
\Leftrightarrow \dfrac{3}{2}\sqrt {x + 2} = \sqrt 2 \\
\Leftrightarrow \sqrt {x + 2} = \dfrac{{2\sqrt 2 }}{3}\\
\Leftrightarrow x + 2 = \dfrac{8}{9}\\
\Leftrightarrow x = \dfrac{8}{9} - 2\\
\Leftrightarrow x = \dfrac{{ - 10}}{9}\left( {tm} \right)\\
Vậy\,x = \dfrac{{ - 10}}{9}
\end{array}$
