Đáp án:
$\begin{array}{l}
2)Dkxd:x \ge 0;x \ne 1\\
a)P = \left( {1 + \dfrac{{x + \sqrt x }}{{\sqrt x + 1}}} \right).\left( {1 - \dfrac{{x - \sqrt x }}{{\sqrt x - 1}}} \right)\\
= \left( {1 + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right).\left( {1 - \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}} \right)\\
= \left( {1 + \sqrt x } \right)\left( {1 - \sqrt x } \right)\\
= 1 - x\\
b)Do:x \ge 0\\
\Leftrightarrow - x \le 0\\
\Leftrightarrow 1 - x \le 1\\
\Leftrightarrow P \le 1\\
\Leftrightarrow GTLN:P = 1\,khi:x = 0
\end{array}$
