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}\)
