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