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