Đáp án:
$\begin{array}{l}
a)P = \dfrac{{1 + x - \sqrt x }}{{\sqrt x - 1}}\left( {x \ge 0;x\# 1} \right)\\
P = \dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{1}{{\sqrt x - 1}} + \sqrt x \\
b)P > 3\\
\Leftrightarrow \dfrac{{1 + x - \sqrt x }}{{\sqrt x - 1}} > 3\\
\Leftrightarrow \dfrac{{1 + x - \sqrt x }}{{\sqrt x - 1}} - 3 > 0\\
\Leftrightarrow \dfrac{{1 + x - \sqrt x - 3\sqrt x + 3}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{x - 4\sqrt x + 4}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x - 1}} > 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x - 2\# 0\\
\sqrt x - 1 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# 4\\
\sqrt x > 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# 4\\
x > 1
\end{array} \right.\\
Vậy\,x > 1;x\# 4
\end{array}$
