Đáp án:
$\begin{array}{l}
Dk:x \ge \frac{1}{4}\\
\sqrt {5x - 1} - \sqrt {4x - 1} \le 3\sqrt x \\
\Rightarrow \sqrt {5x - 1} \le \sqrt {4x - 1} + 3\sqrt x \\
\Rightarrow 5x - 1 \le 4x - 1 + 6\sqrt {4x - 1} .\sqrt x + 9x\\
\Rightarrow - 8x \le 6\sqrt {4{x^2} - x} \\
\Rightarrow - 4x \le 3\sqrt {4{x^2} - x} \\
\Rightarrow 16{x^2} \le 9\left( {4{x^2} - x} \right)\\
\Rightarrow 20{x^2} - 9x \ge 0\\
\Rightarrow x\left( {20x - 9} \right) \ge 0\\
\Rightarrow x \ge \frac{9}{{20}}/x \le 0\\
Vậy\,x \ge \frac{9}{{20}}
\end{array}$
