Đáp án:
$\begin{array}{l}
1)a > 4\\
2)a \in \left\{ {0;1;9;16} \right\}
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
1)Dkxd:a \ge 0;a\# 4\\
P < 1\\
\Leftrightarrow \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}} < 1\\
\Leftrightarrow \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}} - 1 < 0\\
\Leftrightarrow \dfrac{{\sqrt a - 4 - \sqrt a + 2}}{{\sqrt a - 2}} < 0\\
\Leftrightarrow \dfrac{{ - 2}}{{\sqrt a - 2}} < 0\\
\Leftrightarrow \sqrt a - 2 > 0\left( {do: - 2 < 0} \right)\\
\Leftrightarrow \sqrt a > 2\\
\Leftrightarrow a > 4\\
Vậy\,a > 4\\
2)Dkxd:a \ge 0;a\# 4\\
P = \dfrac{{\sqrt a - 4}}{{\sqrt a - 2}} = \dfrac{{\sqrt a - 2 - 2}}{{\sqrt a - 2}}\\
= 1 - \dfrac{2}{{\sqrt a - 2}}\\
P \in Z\\
\Leftrightarrow 2 \vdots \left( {\sqrt a - 2} \right)\\
\Leftrightarrow \left( {\sqrt a - 2} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Leftrightarrow \sqrt a \in \left\{ {0;1;3;4} \right\}\\
\Leftrightarrow a \in \left\{ {0;1;9;16} \right\}\left( {tmdk} \right)\\
Vậy\,a \in \left\{ {0;1;9;16} \right\}
\end{array}$
