Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0\\
P = 1:\left( {\dfrac{{x + 2\sqrt x - 2}}{{x\sqrt x + 1}} - \dfrac{{\sqrt x - 1}}{{x - \sqrt x + 1}} + \dfrac{1}{{\sqrt x + 1}}} \right)\\
= 1:\left( {\dfrac{{x + 2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} - \dfrac{{\sqrt x - 1}}{{x - \sqrt x + 1}} + \dfrac{1}{{\sqrt x + 1}}} \right)\\
= 1:\dfrac{{x + 2\sqrt x - 2 - \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x + 2\sqrt x - 2 - x + 1 + x - \sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x + \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b)x = 7 - 4\sqrt 3 \left( {tmdk} \right)\\
= 4 - 2.2.\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 - \sqrt 3 \\
P = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{7 - 4\sqrt 3 - 2 + \sqrt 3 + 1}}{{2 - \sqrt 3 }}\\
= \dfrac{{6 - 3\sqrt 3 }}{{2 - \sqrt 3 }}\\
= \dfrac{{3\left( {2 - \sqrt 3 } \right)}}{{2 - \sqrt 3 }}\\
= 3\\
c)P = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x - 1 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Leftrightarrow \sqrt x - 1 + \dfrac{1}{{\sqrt x }} \ge 2 - 1\\
\Leftrightarrow P \ge 1\\
\Leftrightarrow GTNN:P = 1\,khi:x = 1\\
d)P = 2\sqrt x - 1\\
\Leftrightarrow \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} = 2\sqrt x - 1\\
\Leftrightarrow x - \sqrt x + 1 = 2x - \sqrt x \\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
