Đáp án:
\(\begin{array}{l}
1)\left[ \begin{array}{l}
x = 3\\
x = - 5\\
x = 1\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.\\
2)\left[ \begin{array}{l}
x = - 1\\
x = - 5\\
x = - 2\\
x = - 4
\end{array} \right.\\
3)\left[ \begin{array}{l}
x = 12\\
x = - 10\\
x = 2\\
x = 0
\end{array} \right.\\
4)\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
5)\left[ \begin{array}{l}
x = 11\\
x = - 21\\
x = 3\\
x = - 13\\
x = - 1\\
x = - 9\\
x = - 3\\
x = - 7\\
x = - 4\\
x = - 6
\end{array} \right.\\
6)\left[ \begin{array}{l}
x = 3\\
x = - 4\\
x = 0\\
x = - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne - 1\\
A = \dfrac{{x + 5}}{{x + 1}} = \dfrac{{x + 1 + 4}}{{x + 1}} = 1 + \dfrac{4}{{x + 1}}\\
A \in Z \to \dfrac{4}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 4\\
x + 1 = - 4\\
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = - 5\\
x = 1\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.\\
2)B = \dfrac{{2x + 4}}{{x + 3}} = \dfrac{{2\left( {x + 3} \right) - 2}}{{x + 3}}\\
= 2 - \dfrac{2}{{x + 3}}\\
B \in Z \to \dfrac{2}{{x + 3}} \in Z \to x + 3 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 2\\
x + 3 = - 2\\
x + 3 = 1\\
x + 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = - 1\\
x = - 5\\
x = - 2\\
x = - 4
\end{array} \right.\\
3)C = \dfrac{{3x + 8}}{{x - 1}} = \dfrac{{3\left( {x - 1} \right) + 11}}{{x - 1}}\\
= 3 + \dfrac{{11}}{{x - 1}}\\
C \in Z \to \dfrac{{11}}{{x - 1}} \in Z\\
\to x - 1 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
x - 1 = 11\\
x - 1 = - 11\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x = - 10\\
x = 2\\
x = 0
\end{array} \right.\\
4)D = \dfrac{{2x - 3}}{{x - 1}} = \dfrac{{2\left( {x - 1} \right) - 1}}{{x - 1}}\\
= 2 - \dfrac{1}{{x - 1}}\\
D \in Z \to \dfrac{1}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
5)E = \dfrac{{5x + 9}}{{x + 5}} = \dfrac{{5\left( {x + 5} \right) - 16}}{{x + 5}}\\
= 5 - \dfrac{{16}}{{x + 5}}\\
E \in Z \to \dfrac{{16}}{{x + 5}} \in Z\\
\to x + 5 \in U\left( {16} \right)\\
\to \left[ \begin{array}{l}
x + 5 = 16\\
x + 5 = - 16\\
x + 5 = 8\\
x + 5 = - 8\\
x + 5 = 4\\
x + 5 = - 4\\
x + 5 = 2\\
x + 5 = - 2\\
x + 5 = 1\\
x + 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = - 21\\
x = 3\\
x = - 13\\
x = - 1\\
x = - 9\\
x = - 3\\
x = - 7\\
x = - 4\\
x = - 6
\end{array} \right.\\
6)F = \dfrac{{2\left( {2x + 1} \right) + 7}}{{2x + 1}} = 2 + \dfrac{7}{{2x + 1}}\\
F \in Z \to \dfrac{7}{{2x + 1}} \in Z\\
\to 2x + 1 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
2x + 1 = 7\\
2x + 1 = - 7\\
2x + 1 = 1\\
2x + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 4\\
x = 0\\
x = - 1
\end{array} \right.
\end{array}\)
