Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{2}{{\sqrt x - 1}} - \dfrac{5}{{x + \sqrt x - 2}}} \right):\left( {1 + \dfrac{{3 - x}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}} \right)\\
= \dfrac{{2\left( {\sqrt x + 2} \right) - 5}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) + 3 - x}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x + 4 - 5}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{x + \sqrt x - 2 + 3 - x}}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}}\\
b)x = 6 - 2\sqrt 5 \left( {tmdk} \right)\\
= {\left( {\sqrt 5 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 - 1\\
P = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{2\sqrt 5 - 2 - 1}}{{\sqrt 5 }} = \dfrac{{2\sqrt 5 - 3}}{{\sqrt 5 }} = \dfrac{{10 - 3\sqrt 5 }}{5}\\
c)P = \dfrac{1}{{\sqrt x }}\left( {x > 0;x \ne 1} \right)\\
\Rightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{1}{{\sqrt x }}\\
\Rightarrow 2x - \sqrt x = \sqrt x + 1\\
\Rightarrow 2x - 2\sqrt x - 1 = 0\\
\Rightarrow x - \sqrt x - \dfrac{1}{2} = 0\\
\Rightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} = \dfrac{3}{4}\\
\Rightarrow \sqrt x = \dfrac{{1 + \sqrt 3 }}{2}\\
\Rightarrow x = \dfrac{{4 + 2\sqrt 3 }}{4} = \dfrac{{2 + \sqrt 3 }}{2}\left( {tmdk} \right)\\
d)P = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{2\sqrt x + 2 - 3}}{{\sqrt x + 1}}\\
= 2 - \dfrac{3}{{\sqrt x + 1}}\\
P \in Z\\
\Rightarrow 3 \vdots \left( {\sqrt x + 1} \right)\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tmdk} \right)\\
x = 4\left( {tmdk} \right)
\end{array} \right.\\
e)P < 1 - \sqrt x \\
\Rightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} < 1 - \sqrt x \\
\Rightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} + \sqrt x - 1 < 0\\
\Rightarrow \dfrac{{2\sqrt x - 1 + x - 1}}{{\sqrt x + 1}} < 0\\
\Rightarrow x + 2\sqrt x - 2 < 0\\
\Rightarrow {\left( {\sqrt x + 1} \right)^2} < 3\\
\Rightarrow \sqrt x < \sqrt 3 - 1\\
\Rightarrow x < 4 - 2\sqrt 3 \\
Vay\,0 \le x < 4 - 2\sqrt 3 \\
f)P = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 1}} = 2 - \dfrac{3}{{\sqrt x + 1}}\\
\sqrt x + 1 \ge 1 \Rightarrow \dfrac{1}{{\sqrt x + 1}} \le 1\\
\Rightarrow \dfrac{3}{{\sqrt x + 1}} \le 3 \Rightarrow 2 - \dfrac{3}{{\sqrt x + 1}} \ge 2 - 3 = - 1\\
\Rightarrow P \ge - 1 \Rightarrow GTNN:P = - 1\,khi:x = 0
\end{array}$
Vậy GTNN của P = -1 khi x=0
