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