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