Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 4;x \ge 0\\
M = \frac{{\sqrt x + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x - 2}}{2}\\
= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
P = A:M = \frac{{\sqrt x - 1}}{{\sqrt x + 2}}.\frac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
= \frac{{\sqrt x - 1}}{{\sqrt x + 1}} = \frac{{\sqrt x + 1 - 2}}{{\sqrt x + 1}} = 1 - \frac{2}{{\sqrt x + 1}}
\end{array}\)
P nguyên ⇔ \(\frac{2}{{\sqrt x + 1}}\) nguyên
\(\begin{array}{l}
\Leftrightarrow \left( {\sqrt x + 1} \right) \in U\left( 2 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = - 2\\
\sqrt x + 1 = 1\\
\sqrt x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = - 3\left( l \right)\\
\sqrt x = 0\\
\sqrt x = - 2\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = 0\left( {tm} \right)
\end{array} \right.
\end{array}\)
