Đáp án:
$\begin{array}{l}
P = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).{\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right)^2}\left( {x \ge 0;x \ne 1} \right)\\
= \left[ {\frac{{\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \frac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right].\frac{{{{\left( {\sqrt x + 1} \right)}^2}.{{\left( {\sqrt x - 1} \right)}^2}}}{2}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}.\left( {\sqrt x - 1} \right)}}.\frac{{{{\left( {\sqrt x + 1} \right)}^2}.{{\left( {\sqrt x - 1} \right)}^2}}}{2}\\
= \frac{{x - \sqrt x - 2 - \left( {x + \sqrt x - 2} \right)}}{2}.\left( {\sqrt x - 1} \right)\\
= \frac{{ - 2\sqrt x \left( {\sqrt x - 1} \right)}}{2}\\
= - x + \sqrt x \\
b)x \ge 0;x \ne 1\\
P = - x + \sqrt x \\
= - \left( {x - 2.\sqrt x .\frac{1}{2} + \frac{1}{4}} \right) + \frac{1}{4}\\
= - {\left( {\sqrt x - \frac{1}{2}} \right)^2} + \frac{1}{4} \le \frac{1}{4}\forall x \le 0;x \ne 1\\
\Rightarrow GTLN:P = \frac{1}{4} \Leftrightarrow x = \frac{1}{4}
\end{array}$
