Đáp án: $a = 1;a = 4$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:a > 0;a\# 9\\
M = \dfrac{{{a^2} - 3a\sqrt a + 2}}{{a - 3\sqrt a }}\\
= \dfrac{{a\left( {a - 3\sqrt a } \right) + 2}}{{a - 3\sqrt a }}\\
= a + \dfrac{2}{{a - 3\sqrt a }}\\
Khi:M \in Z\\
\Leftrightarrow \dfrac{2}{{a - 3\sqrt a }} \in Z\\
\Leftrightarrow a - 3\sqrt a \in \left\{ { - 2; - 1;1;2} \right\}\\
\Leftrightarrow \left[ \begin{array}{l}
a - 3\sqrt a + 2 = 0\\
a - 3\sqrt a + 1 = 0\\
a - 3\sqrt a - 1 = 0\\
a - 3\sqrt a - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right) = 0\\
\sqrt a = \dfrac{{3 \pm \sqrt 5 }}{2}\left( {ktm} \right)\\
\sqrt a = \dfrac{{3 \pm \sqrt {13} }}{2}\left( {ktm} \right)\\
\sqrt a = \dfrac{{3 \pm \sqrt {17} }}{2}\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt a = 1\\
\sqrt a = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = 1\left( {tm} \right)\\
a = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,a = 1;a = 4
\end{array}$
