Đáp án:

 a. \(\dfrac{{x - 8}}{{8x + 16}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x \ne  \pm 2\\
A = \left[ {\dfrac{{x - 2 - 3x - 6 + 3x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{{x - 2}}{8}\\
 = \dfrac{{x - 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x - 2}}{8}\\
 = \dfrac{{x - 8}}{{8x + 16}}\\
b.Thay:x =  - 1\\
 \to A = \dfrac{{ - 1 - 8}}{{8.\left( { - 1} \right) + 16}} = \dfrac{{ - 9}}{8}\\
c.A = \dfrac{{x - 8}}{{8x + 16}} = \dfrac{{x - 8}}{{8\left( {x + 2} \right)}}\\
 \to 8A = \dfrac{{8\left( {x - 8} \right)}}{{8\left( {x + 2} \right)}} = \dfrac{{x - 8}}{{x + 2}}\\
 = \dfrac{{x + 2 - 10}}{{x + 2}} = 1 - \dfrac{{10}}{{x + 2}}\\
Để:A \in Z\\
 \to \dfrac{{10}}{{x + 2}} \in Z\\
 \to x + 2 \in U\left( {10} \right)\\
 \to \left[ \begin{array}{l}
x + 2 = 10\\
x + 2 =  - 10\\
x + 2 = 5\\
x + 2 =  - 5\\
x + 2 = 2\\
x + 2 =  - 2\\
x + 2 = 1\\
x + 2 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 8\\
x =  - 12\\
x = 3\\
x =  - 7\\
x = 0\\
x =  - 4\\
x =  - 1\\
x =  - 3
\end{array} \right.\\
d.\left| {2x - 4} \right| = 5\\
 \to \left[ \begin{array}{l}
2x - 4 = 5\\
2x - 4 =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
A =  - \dfrac{7}{{104}}\\
A =  - \dfrac{{17}}{{24}}
\end{array} \right.\\
e.A =  - 3\\
 \to \dfrac{{x - 8}}{{8x + 16}} =  - 3\\
 \to x - 8 =  - 24x - 48\\
 \to 25x =  - 40\\
 \to x =  - \dfrac{8}{5}\\
g.A > 0\\
 \to \dfrac{{x - 8}}{{8x + 16}} > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 8 > 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 8 < 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x > 8\\
x <  - 2
\end{array} \right.\\
h.A < 0\\
 \to \dfrac{{x - 8}}{{8x + 16}} < 0\\
 \to  - 2 < x < 8\\
i.A >  - \dfrac{3}{2}\\
 \to \dfrac{{x - 8}}{{8x + 16}} >  - \dfrac{3}{2}\\
 \to \dfrac{{2x - 16 + 24x + 48}}{{8\left( {x + 2} \right)}} > 0\\
 \to \dfrac{{26x + 32}}{{8\left( {x + 2} \right)}} > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
26x + 32 > 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
26x + 32 < 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x >  - \dfrac{{16}}{{13}}\\
x <  - 2
\end{array} \right.\\
k.\left| A \right| = 5\\
 \to \left[ \begin{array}{l}
A = 5\\
A =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\dfrac{{x - 8}}{{8x + 16}} = 5\\
\dfrac{{x - 8}}{{8x + 16}} =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x - 8 = 40x + 80\\
x - 8 =  - 40x - 80
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{{88}}{{39}}\\
x =  - \dfrac{{72}}{{41}}
\end{array} \right.
\end{array}\)

Đáp án:

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