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Đáp án:

 a)$P = \dfrac{{4{x^2}}}{{x - 3}}$ với  ${x \ne \left\{ { - 2;0;2;3} \right\}}$

b)$P-49$ khi $x=7$

c)$x \in \left( { - \infty ;3} \right)\backslash \left\{ { - 2;0;2} \right\}$

Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
P = \left( {\dfrac{{2 + x}}{{2 - x}} - \dfrac{{4{x^2}}}{{{x^2} - 4}} - \dfrac{{2 - x}}{{2 + x}}} \right):\dfrac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}\left( {DK:x \ne \left\{ { - 2;0;2;3} \right\}} \right)\\
 = \left( { - \dfrac{{2 + x}}{{x - 2}} - \dfrac{{4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{{x - 2}}{{x + 2}}} \right):\dfrac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
 = \left( {\dfrac{{ - \left( {x + 2} \right)\left( {x + 2} \right) - 4{x^2} + \left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\dfrac{{x\left( {2 - x} \right)}}{{x - 3}}\\
 = \dfrac{{ - 4{x^2} - 8x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x\left( {2 - x} \right)}}{{x - 3}}\\
 = \dfrac{{4x\left( {x + 2} \right)x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{4{x^2}}}{{x - 3}}
\end{array}$

Vậy $P = \dfrac{{4{x^2}}}{{x - 3}}$ với  ${x \ne \left\{ { - 2;0;2;3} \right\}}$

b) Ta có:

$\begin{array}{l}
\left| {x - 5} \right| = 2\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 2\\
x - 5 =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 7\left( c \right)\\
x = 3\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x = 7
\end{array}$

Khi đó:

Với $x=7$ thì $P = \dfrac{{4{x^2}}}{{x - 3}} = \dfrac{{{{4.7}^2}}}{{7 - 3}} = 49$

Vậy $P-49$ khi $x=7$

c) Ta có:

$\begin{array}{l}
\left| P \right| > P\\
 \Leftrightarrow \left| {\dfrac{{4{x^2}}}{{x - 3}}} \right| > \dfrac{{4{x^2}}}{{x - 3}}\\
 \Leftrightarrow \dfrac{{4{x^2}}}{{x - 3}} < 0\\
 \Leftrightarrow x - 3 < 0\\
 \Leftrightarrow x < 3
\end{array}$

Vậy $x \in \left( { - \infty ;3} \right)\backslash \left\{ { - 2;0;2} \right\}$ thỏa mãn đề.