Đáp án:
\(\left[ \begin{array}{l}
x = 36\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
P = \dfrac{{\sqrt x - 6}}{{2\left( {\sqrt x + 2} \right)}}\\
\to 2P = \dfrac{{2\left( {\sqrt x - 6} \right)}}{{2\left( {\sqrt x + 2} \right)}} = \dfrac{{\sqrt x - 6}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 - 8}}{{\sqrt x + 2}} = 1 - \dfrac{8}{{\sqrt x + 2}}\\
P \in Z \to \dfrac{8}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 8\\
\sqrt x + 2 = 4\\
\sqrt x + 2 = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 6\\
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 36\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
