Đáp án:
$\begin{array}{l}
A = \sqrt {x - 2\sqrt x + 1} + \sqrt x \left( {x \ge 0} \right)\\
= \sqrt {{{\left( {\sqrt x - 1} \right)}^2}} + \sqrt x \\
= \left| {\sqrt x - 1} \right| + \sqrt x \\
= \left[ \begin{array}{l}
\sqrt x - 1 + \sqrt x \left( {khi:\sqrt x \ge 1} \right)\\
1 - \sqrt x + \sqrt x \left( {khi:0 \le \sqrt x < 1} \right)
\end{array} \right.\\
= \left[ \begin{array}{l}
2\sqrt x - 1\left( {khi:\sqrt x \ge 1} \right)\\
1\,\left( {khi:0 \le x < 1} \right)
\end{array} \right.\\
Vậy\,A = 2\sqrt x - 1\,hoặc\,A = 1
\end{array}$
