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 $$8 x-8 \sqrt{(x+2)(x-2)}=1 \Rightarrow 8 x-1=8 \sqrt{(x+2)(x-2)}$$ Squaring again gives $$64 x^{2}-16 x+1=64 x^{2}-256$$ so we get that $x=\frac{257}{16}$.