Giúp vs ạ kkk

Điều kiện xác định: \( x > 0; x \neq 4 \) Ta có: \[ \frac{\sqrt{x}}{x - 4} : \frac{1}{\sqrt{4x} + 4} + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{\sqrt{x}}{x - 4} \cdot (\sqrt{4x} + 4) + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{\sqrt{x}}{(x - 4)} \cdot (2\sqrt{x} + 4) + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{\sqrt{x}(2\sqrt{x} + 4)}{(x - 4)} + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{2x + 4\sqrt{x}}{(x - 4)} + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{2(x + 2\sqrt{x})}{(x - 4)} + \frac{4}{2 - \sqrt{x}} \] \[ = \frac{2(x + 2\sqrt{x})}{(x - 4)} + \frac{4}{2 - \sqrt{x}} \cdot \frac{2 + \sqrt{x}}{2 + \sqrt{x}} \] \[ = \frac{2(x + 2\sqrt{x})}{(x - 4)} + \frac{4(2 + \sqrt{x})}{(4 - x)} \] \[ = \frac{2(x + 2\sqrt{x})}{(x - 4)} - \frac{4(2 + \sqrt{x})}{(x - 4)} \] \[ = \frac{2(x + 2\sqrt{x}) - 4(2 + \sqrt{x})}{(x - 4)} \] \[ = \frac{2x + 4\sqrt{x} - 8 - 4\sqrt{x}}{(x - 4)} \] \[ = \frac{2x - 8}{(x - 4)} \] \[ = \frac{2(x - 4)}{(x - 4)} \] \[ = 2 \]