26. Let $x$ and $y$ be the integral and fractional parts of $\sqrt{37-20 \sqrt{3}}$. Find the value of $x+y+\frac{4}{y}$.
(ans. 9.
$$
\begin{array}{l}
\sqrt{37-2(10) \sqrt{3}}=\sqrt{(5-2 \sqrt{3})^{2}}=5-2 \sqrt{3}=1+(4-2 \sqrt{3})=1+2(2- \\
\sqrt{3}) \Rightarrow x=1, y=4-2 \sqrt{3} \text {. Thus, } x+y+\frac{4}{y}=1+4-2 \sqrt{3}+\frac{4}{4-2 \sqrt{3}}= \\
5-2 \sqrt{3}+(4+2 \sqrt{3})=9 \text {. }
\end{array}
$$