Đáp án:

$\begin{array}{l}
1)Dkxd:x \ge 0\\
2.2\sqrt x  - x = 3\\
 \Leftrightarrow x - 4\sqrt x  + 3 = 0\\
 \Leftrightarrow x - \sqrt x  - 3\sqrt x  + 3 = 0\\
 \Leftrightarrow \left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 1\\
\sqrt x  = 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 9
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = 1;x = 9\\
2)a)\left( {\dfrac{1}{{\sqrt 3  - \sqrt 2 }} + \dfrac{{\sqrt 3  - \sqrt 6 }}{{1 - \sqrt 2 }} - \sqrt {27} } \right).\sqrt {5 + 2\sqrt 6 } \\
 = \left( {\dfrac{{\sqrt 3  + \sqrt 2 }}{{3 - 2}} + \dfrac{{\sqrt 3 \left( {1 - \sqrt 2 } \right)}}{{1 - \sqrt 2 }} - 3\sqrt 3 } \right).\sqrt {{{\left( {\sqrt 3  + \sqrt 2 } \right)}^2}} \\
 = \left( {\sqrt 3  + \sqrt 2  + \sqrt 3  - 3\sqrt 3 } \right).\left( {\sqrt 3  + \sqrt 2 } \right)\\
 = \left( {\sqrt 2  - \sqrt 3 } \right).\left( {\sqrt 3  + \sqrt 2 } \right)\\
 = 2 - 3\\
 =  - 1\\
b)\left( {2 + \dfrac{{x + \sqrt x }}{{1 + \sqrt x }}} \right).\left( {2 - \dfrac{{x - \sqrt x }}{{1 - \sqrt x }}} \right)\\
 = \dfrac{{2 + 2\sqrt x  + x + \sqrt x }}{{1 + \sqrt x }}.\dfrac{{2 - 2\sqrt x  - x + \sqrt x }}{{1 - \sqrt x }}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}}{{1 + \sqrt x }}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}{{1 - \sqrt x }}\\
 = \left( {\sqrt x  + 2} \right)\left( {2 + \sqrt x } \right)\\
 = x + 4\sqrt x  + 4
\end{array}$