Giải phương trình: $\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}$.

Điều kiện: x ≥ 6

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

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

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

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

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

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

Þ x – 9 = 0

Û x = 9 (TM)

Vậy x = 9.