Đáp án:
$\begin{array}{l}
a)A = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)P = A - 9\sqrt x \\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} - 9\sqrt x \\
= 1 - \dfrac{1}{{\sqrt x }} - 9\sqrt x \\
= 1 - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right)\\
Do:x > 0;x \ne 1\\
Theo\,Co - si:\\
\dfrac{1}{{\sqrt x }} + 9\sqrt x \ge 2\sqrt {\dfrac{1}{{\sqrt x }}.9\sqrt x } = 6\\
\Leftrightarrow - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right) \le - 6\\
\Leftrightarrow 1 - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right) \le - 5\\
\Leftrightarrow P \le - 5\\
\Leftrightarrow GTLN:P = - 5\\
Khi:\dfrac{1}{{\sqrt x }} = 9\sqrt x \\
\Leftrightarrow x = \dfrac{1}{9}\left( {tmdk} \right)
\end{array}$
