Solution:
We note that
$$\begin{aligned} \frac{1}{4} &= \frac{1}{\sqrt{x}+\sqrt{x-2}}+\frac{1}{\sqrt{x+2}+\sqrt{x}} \\ &= \frac{\sqrt{x}-\sqrt{x-2}}{(\sqrt{x}+\sqrt{x-2})(\sqrt{x}-\sqrt{x-2})} + \frac{\sqrt{x+2}-\sqrt{x}}{(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})} \\ &= \frac{\sqrt{x}-\sqrt{x-2}}{2} + \frac{\sqrt{x+2}-\sqrt{x}}{2} \\ &= \frac{1}{2}(\sqrt{x+2}-\sqrt{x-2}), \end{aligned}$$
so that
$$2 \sqrt{x+2} - 2 \sqrt{x-2} = 1$$
Squaring, we get that
$$8x - 8\sqrt{(x+2)(x-2)} = 1 \Rightarrow 8x - 1 = 8\sqrt{(x+2)(x-2)}.$$
Squaring again gives
$$64x^{2} - 16x + 1 = 64x^{2} - 256$$
so we get that $x = \frac{257}{16}$.