Đáp án:
$\begin{array}{l}
a)2\sqrt {3x - 1} - 4 = 4x\\
\Rightarrow \sqrt {3x - 1} = 2x + 2\left( {dk:x \ge \frac{1}{3}} \right)\\
\Rightarrow 3x - 1 = {\left( {2x + 2} \right)^2}\\
\Rightarrow 3x - 1 = 4{x^2} + 8x + 4\\
\Rightarrow 4{x^2} + 5x + 5 = 0\\
\Rightarrow 4{x^2} + 2.2x.\frac{5}{4} + \frac{{25}}{{16}} + \frac{{55}}{{16}} = 0\\
\Rightarrow {\left( {2x + \frac{5}{4}} \right)^2} + \frac{{55}}{{16}} = 0\left( {vô\,nghiệm} \right)\\
\Rightarrow x \in \emptyset \\
b)\\
\sqrt {x + \sqrt {2x - 1} } + \sqrt {x - \sqrt {2x - 1} } = \sqrt 2 \left( {dkxd:x \ge \frac{1}{2}} \right)\\
\Rightarrow x + \sqrt {2x - 1} + 2\sqrt {x + \sqrt {2x - 1} } .\sqrt {x - \sqrt {2x - 1} } + x - \sqrt {2x - 1} = 2\\
\Rightarrow 2x + 2\sqrt {{x^2} - {{\left( {\sqrt {2x - 1} } \right)}^2}} = 2\\
\Rightarrow \sqrt {{x^2} - 2x + 1} = 1 - x\left( {dk:x \le 1} \right)\\
\Rightarrow {x^2} - 2x + 1 = {x^2} - 2x + 1\left( {\forall x} \right)\\
\Rightarrow \frac{1}{2} \le x \le 1
\end{array}$
