Đáp án:
$\begin{array}{l}
b)B = \left( {\frac{1}{{\sqrt x - 1}} + \frac{{\sqrt x }}{{x - 1}}} \right).\frac{{x - \sqrt x }}{{2\sqrt x + 1}}\\
= \frac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{2\sqrt x + 1}}\\
= \frac{{\sqrt x }}{{\sqrt x + 1}}\\
c)x \ge 0;x \ne 1\\
P = A.B\\
= \frac{{\sqrt x + 1}}{{\sqrt x - 1}}.\frac{{\sqrt x }}{{\sqrt x + 1}}\\
= \frac{{\sqrt x }}{{\sqrt x - 1}}\\
= 1 + \frac{1}{{\sqrt x - 1}}\\
Do:\sqrt x - 1 \ge - 1\\
\Rightarrow \frac{1}{{\sqrt x - 1}} \le - 1\\
\Rightarrow 1 + \frac{1}{{\sqrt x - 1}} \le 0\\
\Rightarrow P \le 0\\
\Rightarrow GTLN:P = 0 \Leftrightarrow x = 0
\end{array}$
