Đáp án:
a. \(\dfrac{{4{x^2}}}{{x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \left\{ { - 2;0;2;3} \right\}\\
P = \left[ {\dfrac{{x + 2}}{{2 - x}} - \dfrac{{4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{{2 - x}}{{2 + x}}} \right]:\dfrac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
= \left[ {\dfrac{{{{\left( {x + 2} \right)}^2} + 4{x^2} - {{\left( {2 - x} \right)}^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}} \right].\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
= \left[ {\dfrac{{{x^2} + 4x + 4 + 4{x^2} - 4 + 4x - {x^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}} \right].\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4{x^2} + 8x}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4{x^2}}}{{x - 3}}\\
b.Để:P \vdots 4\\
\to \dfrac{{4{x^2}}}{{x - 3}} \vdots 4\\
\Leftrightarrow x - 3 \vdots 4\left( {do:4{x^2} \vdots 4\forall x \ne \left\{ { - 2;0;2;3} \right\}} \right)\\
\Leftrightarrow x - 3 \in U\left( 4 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 4\\
x - 3 = - 4\\
x - 3 = 2\\
x - 3 = - 2\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = - 1\\
x = 5\\
x = 1\\
x = 4\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)
