2.110. $\frac{\sqrt{c-d}}{c^{2} \sqrt{2 c}} \cdot\left(\sqrt{\frac{c-d}{c+d}}+\sqrt{\frac{c^{2}+c d}{c^{2}-c d}}\right) ; \quad c=2 ; d=1 / 4$.
Solution.

$$
\begin{aligned}
& \frac{\sqrt{c-d}}{c^{2} \sqrt{2 c}} \cdot\left(\sqrt{\frac{c-d}{c+d}}+\sqrt{\frac{c^{2}+c d}{c^{2}-c d}}\right)=\frac{\sqrt{c-d}}{c^{2} \sqrt{2 c}} \cdot\left(\frac{\sqrt{c-d}}{\sqrt{c+d}}+\sqrt{\frac{c(c+d)}{c(c-d)}}\right)= \\
& =\frac{\sqrt{c-d}}{c^{2} \sqrt{2 c}} \cdot\left(\frac{\sqrt{c-d}}{\sqrt{c+d}}+\frac{\sqrt{c+d}}{\sqrt{c-d}}\right)=\frac{\sqrt{c-d}}{c^{2} \sqrt{2 c}} \cdot \frac{c-d+c+d}{\sqrt{c+d} \cdot \sqrt{c-d}}=\frac{2 c}{c^{2} \cdot \sqrt{2 c} \cdot \sqrt{c+d}}= \\
& =\frac{\sqrt{2}}{c \cdot \sqrt{c} \cdot \sqrt{c+d}}=\frac{\sqrt{2}}{c \sqrt{c^{2}+c d}}=\frac{\sqrt{2}}{2 \sqrt{4+2 \cdot \frac{1}{4}}}=\frac{1}{\sqrt{2} \cdot \sqrt{4+\frac{1}{2}}}=\frac{1}{\sqrt{9}}=\frac{1}{3} .
\end{aligned}
$$

Answer: $\frac{1}{3}$.