117. Simplify the expression:

$$
\sqrt{2+\sqrt{3}}
$$
$\triangle$ Using the formula for the compound square radical, we get:

$$
\sqrt{2+\sqrt{3}}=\sqrt{\frac{2+\sqrt{4-3}}{2}}+\sqrt{\frac{2-\sqrt{4-3}}{2}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{\sqrt{2}(\sqrt{3}+1)}{2}
$$

Answer: $\sqrt{2}(\sqrt{3}+1) / 2$