Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 1;x \ne 3\\
P = \dfrac{{{x^3} - 3}}{{{x^2} - 2x - 3}} - \dfrac{{2\left( {x - 3} \right)}}{{x + 1}} - \dfrac{{x + 3}}{{x - 3}}\\
= \dfrac{{{x^3} - 3 - 2\left( {x - 3} \right)\left( {x - 3} \right) - \left( {x + 3} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
= \dfrac{{{x^3} - 3 - 2{x^2} + 18 - {x^2} - 4x - 3}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
= \dfrac{{{x^3} - 3{x^2} - 4x + 12}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
= \dfrac{{\left( {{x^2} - 4} \right)\left( {x - 3} \right)}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
= \dfrac{{{x^2} - 4}}{{x + 1}}\\
b)P = \dfrac{{{x^2} - 4}}{{x + 1}} = \dfrac{{{x^2} - 1 - 3}}{{x + 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) - 3}}{{x + 1}}\\
= x - 1 - \dfrac{3}{{x + 1}}\\
P \in Z\\
\Rightarrow 3 \vdots \left( {x + 1} \right)\\
\Rightarrow \left( {x + 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Rightarrow x \in \left\{ { - 4; - 2;0;2} \right\}\left( {tmdk} \right)\\
c)x > - 1\\
P = \dfrac{{{x^2} - 4}}{{x + 1}}\\
\Rightarrow P.x + P = {x^2} - 4\\
\Rightarrow {x^2} - P.x - 4 - P = 0\\
\Rightarrow \Delta \ge 0\\
\Rightarrow {P^2} - 4.\left( {4 - P} \right) \ge 0\\
\Rightarrow {P^2} + 4P - 16 \ge 0\\
\Rightarrow {\left( {P + 2} \right)^2} - 20 \ge 0\\
\Rightarrow {\left( {P + 2} \right)^2} \ge 20\\
\Rightarrow P + 2 \ge 2\sqrt 5 \\
\Rightarrow P \ge 2\sqrt 5 - 2\\
\Rightarrow GTNN:P = 2\sqrt 5 - 2\\
Khi:x = 4,06
\end{array}$
