Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{{3x + \sqrt {9x} - 3}}{{x + \sqrt x - 2}} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 2}} - 2} \right):\dfrac{1}{{x - 1}}\\
= \left( {\dfrac{{3x + 3\sqrt x - 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 2}} - 2} \right)\\
.\left( {x - 1} \right)\\
= \dfrac{{3x + 3\sqrt x - 3 + \sqrt x + 2 + \sqrt x - 1 - 2\left( {x + \sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 2} \right)}}.\\
\left( {\sqrt x - 1} \right).\left( {\sqrt x + 1} \right)\\
= \dfrac{{x + 3\sqrt x + 2}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\left( {\sqrt x + 1} \right).\left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}.\left( {\sqrt x + 1} \right)\\
= {\left( {\sqrt x + 1} \right)^2}\\
2)x = 4 - 2\sqrt 3 \left( {tmdk} \right)\\
x = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 3 - 1\\
\Rightarrow P = {\left( {\sqrt 3 - 1 + 1} \right)^2} = {\left( {\sqrt 3 } \right)^2} = 3\\
3)\dfrac{1}{P} = \dfrac{1}{{{{\left( {\sqrt x + 1} \right)}^2}}} \in N\\
\Rightarrow {\left( {\sqrt x + 1} \right)^2} = 1\\
\Rightarrow \sqrt x + 1 = 1\\
\Rightarrow \sqrt x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)
\end{array}$
Vậy x=0 thì 1/P là số tự nhiên bằng 1.
