Đáp án:
g) \(\left[ \begin{array}{l}
x = 5\\
x = - 7\\
x = 2\\
x = - 4\\
x = 1\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{6}{{x - 5}} \in Z\\
\to x - 5 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x - 5 = 6\\
x - 5 = - 6\\
x - 5 = 3\\
x - 5 = - 3\\
x - 5 = 2\\
x - 5 = - 2\\
x - 5 = 1\\
x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = - 1\\
x = 8\\
x = 2\\
x = 7\\
x = 3\\
x = 6\\
x = 4
\end{array} \right.\\
b)\dfrac{{x + 5}}{{x + 1}} = \dfrac{{x + 1 + 4}}{{x + 1}} = 1 + \dfrac{4}{{x + 1}}\\
\dfrac{{x + 5}}{{x + 1}} \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.\\
c)\dfrac{{2x + 1}}{{x - 1}} = \dfrac{{2\left( {x - 1} \right) + 3}}{{x - 1}}\\
= 2 + \dfrac{3}{{x - 1}}\\
\dfrac{{2x + 1}}{{x - 1}} \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)\dfrac{{6x + 5}}{{2x + 1}} = \dfrac{{3\left( {2x + 1} \right) + 2}}{{2x + 1}}\\
= 3 + \dfrac{2}{{2x + 1}}\\
\dfrac{{6x + 5}}{{2x + 1}} \in Z \to \dfrac{2}{{2x + 1}} \in Z\\
\to 2x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
2x + 1 = 2\\
2x + 1 = - 2\\
2x + 1 = 1\\
2x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( l \right)\\
x = - \dfrac{3}{2}\left( l \right)\\
x = 0\\
x = - 1
\end{array} \right.\\
e)A = \dfrac{{3x + 1}}{{2x - 5}}\\
\to 2A = \dfrac{{6x + 2}}{{2x - 5}} = \dfrac{{3\left( {2x - 5} \right) + 17}}{{2x - 5}}\\
= 3 + \dfrac{{17}}{{2x - 5}}\\
A \in Z \to \dfrac{{17}}{{2x - 5}} \in Z\\
\to 2x - 5 \in U\left( {17} \right)\\
\to \left[ \begin{array}{l}
2x - 5 = 17\\
2x - 5 = - 17\\
2x - 5 = 1\\
2x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = - 6\\
x = 3\\
x = 2
\end{array} \right.\\
g)\dfrac{{{x^2} + 5}}{{x + 1}} = \dfrac{{{x^2} - 1 + 6}}{{x + 1}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 6}}{{x + 1}}\\
= \left( {x - 1} \right) + \dfrac{6}{{x + 1}}\\
\dfrac{{{x^2} + 5}}{{x + 1}} \in Z\\
\to \dfrac{6}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 6\\
x + 1 = - 6\\
x + 1 = 3\\
x + 1 = - 3\\
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x = - 7\\
x = 2\\
x = - 4\\
x = 1\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
