Đáp án:
$\begin{array}{l}
P = a.\left\{ {\left( {a - 3} \right) - \left[ {\left( {a + 3} \right) - \left( { - a - 2} \right)} \right]} \right\}\\
= a.\left[ {a - 3 - \left( {a + 3 + a + 2} \right)} \right]\\
= a.\left( {a - 3 - 2a - 5} \right)\\
= a\left( { - a - 8} \right)\\
= - {a^2} - 8a\\
Q = \left[ {a + \left( {a + 3} \right)} \right] - \left[ {\left( {a + 2} \right) - \left( {a - 2} \right)} \right]\\
= \left( {a + a + 3} \right) - \left( {a + 2 - a + 2} \right)\\
= 2a + 3 - 4\\
= 2a - 1\\
Có:Q - P = 2a - 1 - \left( { - {a^2} - 8a} \right)\\
= 2a - 1 + {a^2} + 8a\\
= {a^2} + 10a - 1\\
= {\left( {a + 5} \right)^2} - 26 \ge 0\\
\Rightarrow {\left( {a + 5} \right)^2} \ge 26\\
\Rightarrow \left[ \begin{array}{l}
a \ge \sqrt {26} - 5\\
a \le - \sqrt {26} - 5
\end{array} \right.\\
Vậy\,\left\{ \begin{array}{l}
Q \ge P\, \Leftrightarrow a \in \left( { - \infty ; - \sqrt {26} - 5} \right) \cup \left( {\sqrt {26} - 5; + \infty } \right)\\
Q < P \Leftrightarrow a \in \left( { - \sqrt {26} - 5;\sqrt {26} - 5} \right)
\end{array} \right.
\end{array}$
