Điều kiện: x ≥ 6

\[\sqrt {5{x^2} + 4x} - \sqrt {{x^2} - 3x - 18} = 5\sqrt x \]

\[ \Leftrightarrow \left( {\sqrt {5{x^2} + 4x} - 21} \right) - \left( {\sqrt {{x^2} - 3x - 18} - 6} \right) = 5\sqrt x - 15\]

\[ \Leftrightarrow \frac{{5{x^2} + 4x - 441}}{{\sqrt {5{x^2} + 4x} + 21}} - \frac{{{x^2} - 3x - 18 - 36}}{{\sqrt {{x^2} - 3x - 18} + 6}} = \frac{{25x - 225}}{{5\sqrt x + 15}}\]

\[ \Leftrightarrow \frac{{(x - 9)(5x + 49)}}{{\sqrt {5{x^2} + 4x} + 21}} - \frac{{(x - 9)(x + 6)}}{{\sqrt {{x^2} - 3x - 18} + 6}} - \frac{{25(x - 9)}}{{5\sqrt x + 15}} = 0\]

\[ \Leftrightarrow (x - 9)\left( {\frac{{5x + 49}}{{\sqrt {5{x^2} + 4x} + 21}} - \frac{{x + 6}}{{\sqrt {{x^2} - 3x - 18} + 6}} - \frac{{25}}{{5\sqrt x + 15}}} \right) = 0\]

Dễ thấy \[\frac{{5x + 49}}{{\sqrt {5{x^2} + 4x} + 21}} - \frac{{x + 6}}{{\sqrt {{x^2} - 3x - 18} + 6}} - \frac{{25}}{{5\sqrt x + 15}} > 0\]

Þ x – 9 = 0

Û x = 9 (TM)

Vậy x = 9.