Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
{x^3} - 4x \ne 0\\
x + 2 \ne 0\\
2x - 4 - {x^2} \ne 0\\
2{x^2} + 4x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne \pm 2
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
B = \left( {\dfrac{4}{{{x^3} - 4x}} + \dfrac{1}{{x + 2}}} \right):\dfrac{{2x - 4 - {x^2}}}{{2{x^2} + 4x}}\\
= \left( {\dfrac{4}{{x\left( {{x^2} - 4} \right)}} + \dfrac{1}{{x + 2}}} \right):\dfrac{{2x - 4 - {x^2}}}{{2x\left( {x + 2} \right)}}\\
= \left( {\dfrac{4}{{x\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{1}{{x + 2}}} \right):\dfrac{{2x - 4 - {x^2}}}{{2x\left( {x + 2} \right)}}\\
= \left( {\dfrac{{4 + x.\left( {x - 2} \right)}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}}} \right):\dfrac{{2x - 4 - {x^2}}}{{2x\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 2x + 4}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{2x\left( {x + 2} \right)}}{{ - \left( {{x^2} - 2x + 4} \right)}}\\
= \dfrac{{ - 2}}{{x - 2}}\\
b,\\
x = 1 \Rightarrow B = \dfrac{{ - 2}}{{1 - 2}} = 2\\
c,\\
B \in Z \Leftrightarrow \dfrac{{ - 2}}{{x - 2}} \in Z \Leftrightarrow x - 2 \in \left\{ { - 2;\, - 1;\,1;\,2} \right\}\\
\Rightarrow x \in \left\{ {0;1;3;4} \right\}
\end{array}\)
Kết hợp với ĐKXĐ ta được \(x \in \left\{ {1;3;4} \right\}\)
