Đáp án:
$x=0$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0; x\ne 4$
Ta có:
$\begin{array}{l}
P \in Z\\
\Leftrightarrow \dfrac{{\sqrt x }}{{x - 4}} \in Z\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
\sqrt x \vdots \left( {x - 4} \right)
\end{array} \right.\left( {do:x \in Z} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
{\left( {\sqrt x } \right)^2} \in \left( {x - 4} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
x \vdots \left( {x - 4} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
\left( {\left( {x - 4} \right) + 4} \right) \vdots \left( {x - 4} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
4 \vdots \left( {x - 4} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
\left( {x - 4} \right) \in \left\{ { - 4; - 2; - 1;1;2;4} \right\}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
x \in \left\{ {0;2;3;5;6;8} \right\}
\end{array} \right.\\
\Leftrightarrow x = 0
\end{array}$
Thử lại: $x=0\to P=0$(chọn)
Vậy $x=0$ thỏa mãn đề
