Đáp án:
$\begin{array}{l}
a)P = \left( {\frac{x}{{x - 2}} + \frac{1}{{{x^2} - 4}}} \right):\frac{{x + 1}}{{x + 2}}\\
= \frac{{x\left( {x + 2} \right) + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{x + 2}}{{x + 1}}\\
= \frac{{{x^2} + 2x + 1}}{{x - 2}}.\frac{1}{{x + 1}}\\
= \frac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
= \frac{{x + 1}}{{x - 2}}\\
b)x = \frac{1}{2}\left( {tmdk} \right)\\
\Rightarrow P = \frac{{\frac{1}{2} + 1}}{{\frac{1}{2} - 2}} = \frac{{\frac{3}{2}}}{{ - \frac{3}{2}}} = - 1\\
c)P > 0\\
\Rightarrow \frac{{x + 1}}{{x - 2}} > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 1\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 1\\
x < 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x > 2\\
x < - 1
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x > 2\\
x < - 1;x \ne - 2
\end{array} \right.
\end{array}$
