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