Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
{x^2} - 4 \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 2} \right)\left( {x - 2} \right) \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \ge - 2
\end{array} \right.\\
\Leftrightarrow x \ge 2\\
Vậy\,x \ge 2\\
b)\\
P = 0\\
\Leftrightarrow \sqrt {{x^2} - 4} - 2\sqrt {x + 2} = 0\\
\Leftrightarrow \sqrt {x + 2} .\sqrt {x - 2} - 2\sqrt {x + 2} = 0\\
\Leftrightarrow \sqrt {x + 2} .\left( {\sqrt {x - 2} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
\sqrt {x - 2} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\left( {ktm} \right)\\
x - 2 = 4
\end{array} \right.\\
\Leftrightarrow x = 6\left( {tmdk} \right)\\
Vậy\,x = 6
\end{array}$
