Giải thích các bước giải:
a) Ta có:
$P$ có nghĩa
$ \Leftrightarrow \left\{ \begin{array}{l}
2 - x \ne 0\\
{x^2} - 4 \ne 0\\
2 + x \ne 0\\
{x^2} - 3x \ne 0\\
2{x^2} + {x^3} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne 0\\
x \ne 3
\end{array} \right.$
Vậy $P$ có nghĩa khi $x\not \in \left\{ { \pm 2;0;3} \right\}$
b) Ta có:
$\begin{array}{l}
P = \left( {\dfrac{{2 + x}}{{2 - x}} + \dfrac{{4{x^2}}}{{{x^2} - 4}} - \dfrac{{2 - x}}{{2 + x}}} \right):\dfrac{{{x^2} - 3x}}{{2{x^2} + {x^3}}}\\
= \left( {\dfrac{{2 + x}}{{2 - x}} + \dfrac{{4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{{x - 2}}{{x + 2}}} \right).\dfrac{{{x^2}\left( {x + 2} \right)}}{{x\left( {x - 3} \right)}}\\
= \left( {\dfrac{{ - \left( {2 + x} \right)\left( {x + 2} \right) + 4{x^2} + {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\dfrac{{{x^2}\left( {x + 2} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4{x^2} - 8x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{{x^2}\left( {x + 2} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{{x^2}\left( {x + 2} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4{x^2}}}{{x - 3}}
\end{array}$
Vậy $P = \dfrac{{4{x^2}}}{{x - 3}}$ với $x\not \in \left\{ { \pm 2;0;3} \right\}$
