Đáp án:
b. \(\left[ \begin{array}{l}
x = - 1\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
c.M = \frac{{x - 6}}{{2x - 1}}\\
\to 2M = \frac{{2x - 12}}{{2x - 1}} = \frac{{2x - 1 - 11}}{{2x - 1}}\\
= 1 - \frac{{11}}{{2x - 1}}\\
Để:M \in Z\\
\to \frac{{11}}{{2x - 1}} \in Z\\
\to 2x - 1 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
2x - 1 = 11\\
2x - 1 = - 11\\
2x - 1 = 1\\
2x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 12\\
2x = - 10\\
2x = 2\\
2x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - 5\\
x = 1\\
x = 0
\end{array} \right.\\
d.B = \frac{{x + 2 - 1}}{{x + 2}} = 1 - \frac{1}{{x + 2}}\\
Để:B \in Z\\
\to \frac{1}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
