Đáp án:
$\begin{array}{l}
\left| {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right|\\
= \sqrt {{{\left( {\left| {{x_1}} \right| + \left| {{x_2}} \right|} \right)}^2}} \\
= \sqrt {x_1^2 + x_2^2 + 2\left| {{x_1}{x_2}} \right|} \\
= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2} + 2\left| {{x_1}{x_2}} \right|} \\
b)\\
\sqrt {{x_1}} = \sqrt {{x_2}} \\
\Rightarrow \left\{ \begin{array}{l}
{x_1};{x_2} \ge 0\\
{x_1} = {x_2}
\end{array} \right.
\end{array}$
