Đáp án:
$\begin{array}{l}
B = \sqrt {x - 2 + 2\sqrt {x - 3} } \left( {x \ge 3} \right)\\
= \sqrt {x - 3 + 2\sqrt {x - 3} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 3} + 1} \right)}^2}} \\
= \sqrt {x - 3} + 1\\
C = \sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } \left( {x \ge 1} \right)\\
= \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right|\\
+ Khi:\sqrt {x - 1} - 1 \ge 0\\
\Rightarrow x \ge 2\\
\Rightarrow C = \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1\\
= 2\sqrt {x - 1} \\
+ Khi:\sqrt {x - 1} - 1 < 0\\
\Rightarrow 1 \le x < 2\\
\Rightarrow C = \sqrt {x - 1} + 1 + 1 - \sqrt {x - 1} \\
= 2
\end{array}$
