Đáp án:
$\begin{array}{l}
{\rm{a)2x + 8}}\sqrt {2x - 1} = 21\left( {dk:x \ge \frac{1}{2}} \right)\\
\Rightarrow 2x - 1 + 8\sqrt {2x - 1} - 20 = 0\\
\Rightarrow {\left( {\sqrt {2x - 1} } \right)^2} + 8\sqrt {2x - 1} - 20 = 0\\
\Rightarrow \left( {\sqrt {2x - 1} - 2} \right)\left( {\sqrt {2x - 1} + 10} \right) = 0\\
\Rightarrow \sqrt {2x - 1} = 2\left( {do:\sqrt {2x - 1} \ge 0} \right)\\
\Rightarrow 2x - 1 = 4\\
\Rightarrow x = \frac{5}{2}\left( {tm} \right)\\
b)\sqrt {x - 2011} + \sqrt {4x - 8044} = 3\left( {dk:x \ge 2011} \right)\\
\Rightarrow \sqrt {x - 2011} + 2\sqrt {x - 2011} = 3\\
\Rightarrow \sqrt {x - 2011} = 1\\
\Rightarrow x - 2011 = 1\\
\Rightarrow x = 2012\left( {tm} \right)
\end{array}$
