Đáp án:
\(\left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{5x}}{{{x^2} - 4}} = \dfrac{{5\left( {x - 2} \right) + 10}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{5}{{x + 2}} + \dfrac{{10}}{{{x^2} - 4}}\\
A \in Z \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{5}{{x + 2}} \in Z\\
\dfrac{{10}}{{{x^2} - 4}} \in Z
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 2 \in U\left( 5 \right)\\
{x^2} - 4 \in U\left( {10} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x + 2 = 5\\
x + 2 = - 5\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\left[ \begin{array}{l}
{x^2} - 4 = 10\\
{x^2} - 4 = - 10\left( l \right)\\
{x^2} - 4 = 5\\
{x^2} - 4 = - 5\left( l \right)\\
{x^2} - 4 = 2\\
{x^2} - 4 = - 2\\
{x^2} - 4 = 1\\
{x^2} - 4 = - 1
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 3\\
x = - 7\\
x = - 1\\
x = - 3
\end{array} \right.\\
\left[ \begin{array}{l}
x = \pm \sqrt {14} \\
x = 3\\
x = - 3\\
x = \pm \sqrt 6 \\
x = \pm \sqrt 2 \\
x = \pm \sqrt 5 \\
x = \pm \sqrt 3
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.
\end{array}\)
