Сократите дробь. Спасибо огромное!!!

Решите задачу:

$\frac{x+3\sqrt{5x}}{x-45}=\frac{\sqrt{x}\cdot (\sqrt{x}+3\sqrt5)}{(\sqrt{x})^2-(3\sqrt5)^2}=\frac{\sqrt{x}\cdot (\sqrt{x}+3\sqrt5)}{(\sqrt{x}-3\sqrt5)(\srqt{x}+3\sqrt5)}=\frac{\sqrt{x}}{\sqrt{x}-3\sqrt5}\\\\\\\frac{\sqrt{125a^3}+\sqrt{64b^3}}{5a-2\sqrt{5ab}+4b}=\frac{(\sqrt{5a})^3+(\sqrt{4b})^3}{(\sqrt{5a})^2-2\cdot \sqrt{5a}\cdot \sqrt{b}+(2\sqrt{b})^2}=\frac{(\sqrt{5a}+\sqrt{4b})(5a-\sqrt{5a}\cdot \sqrt{4b}+4b)}{5a-\sqrt{5a}\cdot \sqrt{4b}+4b}=\\\\=\sqrt{5a}+\sqrt{4b}=\sqrt{5a}+2\sqrt{b}$