\[\begin{array}{l}P = a - \left\{ {\left[ {a - 3} \right] - \left[ {\left[ {a + 3} \right] - \left[ { - a - 2} \right]} \right]} \right\}\\\,\,\,\, = a - \left[ {\left[ {a - 3} \right] - \left[ {a + 3 + a + 2} \right]} \right]\\\,\,\,\, = a - \left[ {a - 3 - a - 3 - a - 2} \right]\\\,\,\,\, = a - \left[ { - a - 8} \right]\\\,\,\,\, = a + a + 8\\\,\,\,\, = 2a + 8\end{array}\]
\[\begin{array}{l}Q = \left[ {a + \left[ {a + 3} \right]} \right] + \left[ {\left[ {a + 2} \right] - \left[ {a - 2} \right]} \right]\\\,\,\,\,\, = \left[ {a + a + 3} \right] + \left[ {a + 2 - a + 2} \right]\\\,\,\,\, = \left[ {2a + 3} \right] + 4\\\,\,\,\, = 2a + 3 + 4\\\,\,\,\, = 2a + 7\end{array}\]
\[p=a-\left\{\left[a-3\right]-\left[\left[a+3\right]-\left[-a-2\right]\right]\right\}\]
\[=a-\left\{a-3-\left[a+3+a+2\right]\right\}\]
\[=a-\left\{a-3-a-3-a-2\right\}\]
\[=a-\left\{-a-8\right\}\]
\[=a+a+8\]
\[=2a+8\]
\[Q=\left[a+\left[a+3\right]\right]-\left[\left[a+2\right]-\left[a-2\right]\right]\]
\[=\left[a+a+3\right]-\left[a+2-a+2\right]\]
\[=2a+3-4\]
\[=2a-1\]
Xét hiệu \[P-Q=\left[2a+8\right]-\left[2a-1\right]\]
\[=2a+8-2a+1\]
\[=9>0\]
Vậy: \[P>Q\]
Xét \[P - 2 = \dfrac{{2\sqrt x }}{{\sqrt x + 1}} - 2 = \dfrac{{2\sqrt x - 2\left[ {\sqrt x + 1} \right]}}{{\sqrt x + 1}}\]
\[P - 2 = \dfrac{{2\sqrt x - 2\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{ - 2}}{{\sqrt x + 1}}\]
Ta có \[x \ge 0 \Rightarrow \sqrt x \ge 0 \Rightarrow \sqrt x + 1 \ge 1 \Rightarrow \sqrt x + 1 > 0\]
\[ \Rightarrow \dfrac{{ - 2}}{{\sqrt x + 1}} < 0 \Leftrightarrow P - 2 < 0 \Leftrightarrow P < 2\,\,\forall x\,\,tm\,\left\{ \begin{array}{l}x \ge 0\\x \ne 1\end{array} \right.\].