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