Đáp án:
a. \(\dfrac{{x + 2003}}{x}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \left( {\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} + \dfrac{{{x^2} - 4x - 1}}{{{x^2} - 1}}} \right).\dfrac{{x + 2003}}{x}\\
= \left[ {\dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1 + {x^2} - 4x - 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}} \right].\dfrac{{x + 2003}}{x}\\
= \dfrac{{{x^2} - 1}}{{{x^2} - 1}}.\dfrac{{x + 2003}}{x}\\
= \dfrac{{x + 2003}}{x}\\
b.A = \dfrac{{x + 2003}}{x} = 1 + \dfrac{{2003}}{x}\\
A \in Z\\
\Leftrightarrow \dfrac{{2003}}{x} \in Z\\
\Leftrightarrow x \in U\left( {2003} \right)\\
\to \left[ \begin{array}{l}
x = 2003\\
x = - 2003\\
x = 1\\
x = - 1
\end{array} \right.
\end{array}\)
