Đáp án:
b. \(\left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne - 5\\
A = \dfrac{{x - 2}}{{x + 5}} = \dfrac{{x + 5 - 7}}{{x + 5}} = 1 - \dfrac{7}{{x + 5}}\\
A \in Z\\
\Leftrightarrow \dfrac{7}{{x + 5}} \in Z\\
\Leftrightarrow x + 5 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x + 5 = 7\\
x + 5 = - 7\\
x + 5 = 1\\
x + 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 12\\
x = - 4\\
x = - 6
\end{array} \right.\\
b.DK:x \ne - 1\\
B = \dfrac{{2x - 1}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) - 3}}{{x + 1}} = 2 - \dfrac{3}{{x + 1}}\\
B \in Z \Leftrightarrow \dfrac{3}{{x + 1}} \in Z\\
\Leftrightarrow x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = - 3\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 4\\
x = 0\\
x = - 2
\end{array} \right.\\
c.DK:x \ne \dfrac{1}{3}\\
C = \dfrac{{4 - x}}{{3x - 1}}\\
\to 3C = \dfrac{{12 - 3x}}{{3x - 1}} = \dfrac{{ - \left( {3x - 1} \right) + 11}}{{3x - 1}} = - 1 + \dfrac{{11}}{{3x - 1}}\\
C \in Z \Leftrightarrow \dfrac{{11}}{{3x - 1}} \in Z\\
\Leftrightarrow 3x - 1 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
3x - 1 = 11\\
3x - 1 = - 11\\
3x - 1 = 1\\
3x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - \dfrac{{10}}{3}\left( l \right)\\
x = \dfrac{2}{3}\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)
