Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
b)A = \dfrac{{x + 1 - 2\sqrt x }}{{\sqrt x - 1}} + \dfrac{{x + \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x - 1}} + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}} + \sqrt x \\
= \sqrt x - 1 + \sqrt x \\
= 2\sqrt x - 1
\end{array}$
