Đáp án:
d) \(\left[ \begin{array}{l}
x = 14\\
x = - 8\\
x = 4\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B7.1\\
a) - \dfrac{3}{{x - 1}} \in Z\\
\Leftrightarrow x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - 2\\
x = 2\\
x = 0
\end{array} \right.\\
b) - \dfrac{4}{{2x - 1}} \in Z\\
\to 2x - 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
2x - 1 = 4\\
2x - 1 = - 4\\
2x - 1 = 2\\
2x - 1 = - 2\\
2x - 1 = 1\\
2x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( l \right)\\
x = - \dfrac{3}{2}\left( l \right)\\
x = \dfrac{3}{2}\left( l \right)\\
x = - \dfrac{1}{2}\left( l \right)\\
x = 1\\
x = 0
\end{array} \right.\\
c)\dfrac{{3x + 7}}{{x - 1}} = \dfrac{{3\left( {x - 1} \right) + 10}}{{x - 1}} = 3 + \dfrac{{10}}{{x - 1}}\\
\dfrac{{3x + 7}}{{x - 1}} \in Z\\
\to \dfrac{{10}}{{x - 1}} \in Z\\
\to x - 1 \in U\left( {10} \right)\\
\to \left[ \begin{array}{l}
x - 1 = 10\\
x - 1 = - 10\\
x - 1 = 5\\
x - 1 = - 5\\
x - 1 = 2\\
x - 1 = - 2\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = - 9\\
x = 6\\
x = - 4\\
x = 3\\
x = - 1\\
x = 2\\
x = 0
\end{array} \right.\\
d)\dfrac{{4x - 1}}{{3 - x}} = - \dfrac{{4\left( {x - 3} \right) + 11}}{{x - 3}} = - 4 - \dfrac{{11}}{{x - 3}}\\
\dfrac{{4x - 1}}{{3 - x}} \in Z \Leftrightarrow \dfrac{{11}}{{x - 3}} \in Z\\
\Leftrightarrow x - 3 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
x - 3 = 11\\
x - 3 = - 11\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 14\\
x = - 8\\
x = 4\\
x = 2
\end{array} \right.
\end{array}\)
