5. $[\mathbf{5}]$ Find all real values of $x$ for which
$$
\frac{1}{\sqrt{x}+\sqrt{x-2}}+\frac{1}{\sqrt{x+2}+\sqrt{x}}=\frac{1}{4} .
$$
Answer: $\frac{257}{16}$ 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 $x=\frac{257}{16}$.