Đáp án:
$\begin{array}{l}
Dkxd:x > 0;x\# 1\\
a)A = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)x = \dfrac{1}{{7 - 4\sqrt 3 }}\left( {tmdk} \right)\\
= \dfrac{{7 + 4\sqrt 3 }}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}}\\
= \dfrac{{7 + 4\sqrt 3 }}{1}\\
= 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = 2 + \sqrt 3 \\
B = 9\sqrt x = 9.\left( {2 + \sqrt 3 } \right) = 18 + 9\sqrt 3 \\
c)P = A - B\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} - 9\sqrt x \\
= \dfrac{{\sqrt x - 1 - 9x}}{{\sqrt x }}\\
= \dfrac{{ - \left( {9x - \sqrt x } \right) - 1}}{{\sqrt x }}\\
= \dfrac{{ - \left( {9x - 2.3\sqrt x .\dfrac{1}{6} + \dfrac{1}{{36}}} \right) + \dfrac{1}{{36}} - 1}}{{\sqrt x }}\\
= \dfrac{{ - {{\left( {3\sqrt x - \dfrac{1}{6}} \right)}^2} - \dfrac{{35}}{{36}}}}{{\sqrt x }} \le \dfrac{{ - 35}}{{36}}\\
\Leftrightarrow P \le \dfrac{{ - 35}}{{36}}\\
\Leftrightarrow GTLN:P = \dfrac{{ - 35}}{{36}}\,khi:\sqrt x = \dfrac{1}{{18}} \Leftrightarrow x = \dfrac{1}{{324}}
\end{array}$
