Đáp án:
a) \(\left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 + 1}}{{\sqrt x - 2}} = 1 + \dfrac{1}{{\sqrt x - 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \sqrt x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\\
b)P = 1 + \dfrac{1}{{\sqrt x - 2}} \le \dfrac{1}{2}\\
Mà:P \ge 0\forall x \ge 0;x \ne 4\\
\to 0 \le x < \dfrac{1}{2}\\
P \in Z\\
\Leftrightarrow P = 0\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} = 0\\
\to \sqrt x - 1 = 0\\
\to x = 1
\end{array}\)
