Đáp án:
$\begin{array}{l}
1)x > 0;x\# 1\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{x\sqrt x - 1}}} \right).\dfrac{{3\sqrt x - 3}}{{x + \sqrt x }}\\
= \dfrac{{x + \sqrt x + 1 - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{3\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}.\dfrac{3}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{3}{{x + \sqrt x + 1}}\\
2)x = 4 + 2\sqrt 3 \left( {tmdk} \right)\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 + 1\\
P = \dfrac{3}{{4 + 2\sqrt 3 + \sqrt 3 + 1 + 1}}\\
= \dfrac{3}{{6 + 3\sqrt 3 }} = \dfrac{1}{{2 + \sqrt 3 }} = \dfrac{{2 - \sqrt 3 }}{{4 - 3}}\\
= 2 - \sqrt 3
\end{array}$
