Đáp án:
$\begin{array}{l}
2)Dkxd:\left\{ \begin{array}{l}
2 - x \ge 0\\
{x^2} - 4 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
\left[ \begin{array}{l}
x \le - 2\\
x \ge 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x \le - 2
\end{array} \right.\\
\sqrt {2 - x} - \sqrt {{x^2} - 4} = 0\\
\Leftrightarrow \sqrt {2 - x} - \sqrt {2 - x} .\sqrt { - x - 2} = 0\\
\Leftrightarrow \sqrt {2 - x} .\left( {1 - \sqrt { - x - 2} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
\sqrt { - x - 2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
- x - 2 = 1 \Leftrightarrow x = - 3\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 2;x = - 3\\
4)Dkxd:x \ge 2\\
\sqrt {x - 2} + \sqrt {4x - 8} - \dfrac{1}{2}\sqrt {9x - 18} = 2\\
\Leftrightarrow \sqrt {x - 2} + 2\sqrt {x - 2} - \dfrac{1}{2}.3\sqrt {x - 2} = 2\\
\Leftrightarrow \dfrac{3}{2}\sqrt {x - 2} = 2\\
\Leftrightarrow \sqrt {x - 2} = \dfrac{4}{3}\\
\Leftrightarrow x - 2 = \dfrac{{16}}{9}\\
\Leftrightarrow x = \dfrac{{34}}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{34}}{9}\\
6)Dkxd:x \ge 2\\
\sqrt {x + 1} = 2 + \sqrt {x - 2} \\
\Leftrightarrow x + 1 = 4 + 4\sqrt {x - 2} + x - 2\\
\Leftrightarrow 4\sqrt {x - 2} = - 1\\
\Leftrightarrow \sqrt {x - 2} = - \dfrac{1}{4}\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$
