Đáp án:
\(\dfrac{1}{4} \le x \le 1\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0\\
P = \dfrac{{1 - \sqrt x }}{{\sqrt x }}\\
P < \sqrt P \\
\to \sqrt P .\left( {\sqrt P - 1} \right) < 0\\
\to \sqrt P - 1 < 0\left( {do:\sqrt P \ge 0\forall P \ge 0} \right)\\
\to \left\{ \begin{array}{l}
\dfrac{{1 - \sqrt x }}{{\sqrt x }} \ge 0\\
\sqrt {\dfrac{{1 - \sqrt x }}{{\sqrt x }}} - 1 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 - \sqrt x \ge 0\left( {Do:\sqrt x > 0\forall x > 0} \right)\\
\dfrac{{1 - \sqrt x }}{{\sqrt x }} < 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 \ge x\\
\dfrac{{1 - \sqrt x - \sqrt x }}{{\sqrt x }} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 \ge x\\
1 - 2\sqrt x < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 \ge x\\
\dfrac{1}{4} \le x
\end{array} \right.\\
\to \dfrac{1}{4} \le x \le 1
\end{array}\)
