Đáp án:
$\begin{array}{l}
a)\dfrac{x}{{\sqrt 2 }} + \sqrt 2 = 3\\
\Leftrightarrow \dfrac{x}{{\sqrt 2 }} = 3 - \sqrt 2 \\
\Leftrightarrow x = 3\sqrt 2 - 2\\
Vậy\,x = 3\sqrt 2 - 2\\
b)3\sqrt 2 .x - 2\sqrt 2 = \sqrt 6 - \sqrt 3 x\\
\Leftrightarrow 3\sqrt 2 .x + \sqrt 3 x = 2\sqrt 2 + \sqrt 6 \\
\Leftrightarrow x.\left( {3\sqrt 2 + \sqrt 3 } \right) = 2\sqrt 2 + \sqrt 6 \\
\Leftrightarrow x = \dfrac{{2\sqrt 2 + \sqrt 6 }}{{3\sqrt 2 + \sqrt 3 }}\\
= \dfrac{{\left( {2\sqrt 2 + \sqrt 6 } \right)\left( {\sqrt 6 - 1} \right)}}{{\sqrt 3 \left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 - 1} \right)}}\\
= \dfrac{{4\sqrt 3 - 2\sqrt 2 + 6 - \sqrt 6 }}{{5\sqrt 3 }}\\
Vậy\,x = \dfrac{{4\sqrt 3 - 2\sqrt 2 + 6 - \sqrt 6 }}{{5\sqrt 3 }}
\end{array}$
