Đáp án:
d) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a)DK:x \ne \left\{ { - 2;0;2} \right\}\\
b)A = \left[ {\dfrac{{{x^2}}}{{x\left( {{x^2} - 4} \right)}} - \dfrac{6}{{3\left( {x - 2} \right)}} + \dfrac{1}{{x + 2}}} \right]:\dfrac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}\\
= \left[ {\dfrac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{2}{{x - 2}} + \dfrac{1}{{x + 2}}} \right]:\dfrac{6}{{x + 2}}\\
= \dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{6}{{x + 2}}\\
= \dfrac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{6}{{x + 2}}\\
= \dfrac{{ - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{6}{{x + 2}}\\
= \dfrac{{ - 1}}{{x - 2}}\\
c)Thay:x = 2009\\
\to A = \dfrac{{ - 1}}{{2009 - 2}} = - \dfrac{1}{{2007}}\\
d)A \in Z \Leftrightarrow \dfrac{{ - 1}}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
