Đáp án:
$\begin{array}{l}
b)\sqrt {x - 2 + 2\sqrt {x - 3} } \left( {x \ge 3} \right)\\
= \sqrt {x - 3 + 2\sqrt {x - 3} + 1} \left( {x \ge 3} \right)\\
= \sqrt {{{\left( {\sqrt {x - 3} + 1} \right)}^2}} \left( {x \ge 3} \right)\\
= \left| {\sqrt {x - 3} + 1} \right|\\
= \sqrt {x - 3} + 1\\
c)\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } \left( {x \ge 1} \right)\\
= \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right|\\
d)x \ge 0\\
\sqrt {x - 2\sqrt x + 1} + \sqrt {x + 2\sqrt x + 1} \\
= \sqrt {{{\left( {\sqrt x - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} \\
= \left| {\sqrt x - 1} \right| + \sqrt x + 1
\end{array}$
