Đáp án:
$\begin{array}{l}
a)P = \left( {\frac{{a\sqrt a - 1}}{{a - \sqrt a }} - \frac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right):\frac{{a + 2}}{{a - 2}}\\
= \left( {\frac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \frac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right).\frac{{a - 2}}{{a + 2}}\\
= \frac{{a + \sqrt a + 1 - \left( {a - \sqrt a + 1} \right)}}{{\sqrt a }}.\frac{{a - 2}}{{a + 2}}\\
= \frac{{2\sqrt a }}{{\sqrt a }}.\frac{{a - 2}}{{a + 2}}\\
= \frac{{2a - 4}}{{a + 2}}\\
b)a > 0;a \ne 1;a \ne 2\\
P \in Z\\
\Rightarrow \frac{{2a - 4}}{{a + 2}} \in Z\\
\Rightarrow \frac{{2\left( {a + 2} \right) - 8}}{{a + 2}} \in Z\\
\Rightarrow 2 - \frac{8}{{a + 2}} \in Z\\
\Rightarrow \left( {a + 2} \right) \in U\left( 8 \right)\\
Do:a > 0;a \ne 1;a \ne 2\\
\Rightarrow a + 2 > 2;\left( {a + 2} \right) \ne 3;\left( {a + 2} \right) \ne 4\\
\Rightarrow \left( {a + 2} \right) = 8\\
\Rightarrow a = 6
\end{array}$
Vậy a=6
