Đáp án:
$x \in \left\{ {0;1} \right\}$
Giải thích các bước giải:
ĐK: $x\ge 0$
Ta có:
Với $x\in Z$ thì $P \in Z$
$\begin{array}{l}
\Leftrightarrow \left( {\sqrt x - 1} \right) \vdots \left( {x + \sqrt x + 1} \right)\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) \vdots \left( {x + \sqrt x + 1} \right)\\
\Leftrightarrow \left( {x + \sqrt x - 2} \right) \vdots \left( {x + \sqrt x + 1} \right)\\
\Leftrightarrow \left( {x + \sqrt x + 1 - 3} \right) \vdots \left( {x + \sqrt x + 1} \right)\\
\Leftrightarrow 3 \vdots \left( {x + \sqrt x + 1} \right)\\
\Leftrightarrow \left( {x + \sqrt x + 1} \right) \in U\left( 3 \right) = \left\{ { \pm 1; \pm 3} \right\} (1)
\end{array}$
Mà lại có:
$x + \sqrt x + 1 = {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 1,\forall x \ge 0$
Nên
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
x + \sqrt x + 1 = 1\\
x + \sqrt x + 1 = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + \sqrt x = 0\\
x + \sqrt x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x \left( {\sqrt x + 1} \right) = 0\\
\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
\sqrt x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}$
Thử lại:
$\begin{array}{l}
x = 0 \Rightarrow P = \dfrac{{ - 1}}{1} = - 1 \in Z \Rightarrow x = 0\left( c \right)\\
x = 1 \Rightarrow P = \dfrac{0}{3} = 0 \in Z \Rightarrow x = 1\left( c \right)
\end{array}$
Vậy $x \in \left\{ {0;1} \right\}$ thỏa mãn đề.
