Đáp án:
\[\left[ \begin{array}{l}
a = 2\\
a = 0
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \frac{{{a^2} + 4a + 4}}{{{a^3} + 2{a^2} - 4a - 8}} = \frac{{{{\left( {a + 2} \right)}^2}}}{{{a^2}\left( {a + 2} \right) - 4\left( {a + 2} \right)}} = \frac{{{{\left( {a + 2} \right)}^2}}}{{\left( {{a^2} - 4} \right)\left( {a + 2} \right)}} = \frac{{{{\left( {a + 2} \right)}^2}}}{{\left( {a - 2} \right){{\left( {a + 2} \right)}^2}}} = \frac{1}{{a - 2}}\\
P \in Z \Leftrightarrow \frac{1}{{a - 2}} \in Z \Rightarrow \left[ \begin{array}{l}
a - 1 = 1\\
a - 1 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 2\\
a = 0
\end{array} \right.
\end{array}\)
