Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 3;x \ne - 3\\
P = \left( {\dfrac{3}{{x - 3}} - \dfrac{3}{{x + 3}}} \right).\dfrac{{{x^2} + 6x + 9}}{{18}}\\
= \dfrac{{3\left( {x + 3} \right) - 3\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{{{\left( {x + 3} \right)}^2}}}{{18}}\\
= \dfrac{{3x + 9 - 3x + 9}}{{x - 3}}.\dfrac{{x + 3}}{{18}}\\
= \dfrac{{x + 3}}{{x - 3}}\\
b)\left| {x - 2} \right| = 1\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 1 \Leftrightarrow x = 3\left( {ktm} \right)\\
x - 2 = - 1 \Leftrightarrow x = 1\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = 1\\
\Leftrightarrow P = \dfrac{{x + 3}}{{x - 3}} = \dfrac{{1 + 3}}{{1 - 3}} = - 2\\
c)P < 1\\
\Leftrightarrow \dfrac{{x + 3}}{{x - 3}} < 1\\
\Leftrightarrow \dfrac{{x + 3}}{{x - 3}} - 1 < 0\\
\Leftrightarrow \dfrac{{x + 3 - x + 3}}{{x - 3}} < 0\\
\Leftrightarrow \dfrac{6}{{x - 3}} < 0\\
\Leftrightarrow x - 3 < 0\\
\Leftrightarrow x < 3\\
Vậy\,x < 3;x \ne - 3\\
d)P = \dfrac{{x + 3}}{{x - 3}} = \dfrac{{x - 3 + 6}}{{x - 3}} = 1 + \dfrac{6}{{x - 3}}\\
P \in Z\\
\Leftrightarrow \dfrac{6}{{x - 3}} \in Z\\
\Leftrightarrow \left( {x - 3} \right) \in \left\{ { - 6; - 3; - 2; - 1;1;2;3;6} \right\}\\
\Leftrightarrow x \in \left\{ { - 3;0;1;2;4;5;6;9} \right\}\\
Do:x \ne - 3\\
\Leftrightarrow x \in \left\{ {0;1;2;4;5;6;9} \right\}\\
Vậy\,x \in \left\{ {0;1;2;4;5;6;9} \right\}
\end{array}$
