Đáp án:
d) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
A = \dfrac{{4\left( {x - 2} \right) + 2\left( {x + 2} \right) + 6 - 5x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x - 8 + 2x + 4 + 6 - 5x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x - 2}}\\
b)A = 1\\
\to \dfrac{1}{{x - 2}} = 1\\
\to x - 2 = 1\\
\to x = 3\\
c)A > 1\\
\to \dfrac{1}{{x - 2}} > 1\\
\to \dfrac{{1 - x + 2}}{{x - 2}} > 0\\
\to \dfrac{{3 - x}}{{x - 2}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3 - x > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3 - x < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
3 > x > 2\\
\left\{ \begin{array}{l}
x > 3\\
x < 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
d)A \in Z \Leftrightarrow \dfrac{1}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
