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