2.131. $\left(\frac{4}{3-\sqrt{5}}\right)^{2}-\left(\frac{6-5 \sqrt{6}}{5-\sqrt{6}}\right)^{2}=2 \sqrt{61+24 \sqrt{5}}$.
## Solution.

\[
\begin{aligned}
& \frac{16}{9-6 \sqrt{5}+5}-\left(-\frac{\sqrt{6}(\sqrt{6}-5)}{\sqrt{6}-5}\right)^{2}=2 \sqrt{61+24 \sqrt{5}} \\
& \frac{16}{14-6 \sqrt{5}}-6=2 \sqrt{61+24 \sqrt{5}}, \quad \frac{8}{7-3 \sqrt{5}}-6=2 \sqrt{61+24 \sqrt{5}} \\
& \frac{4}{7-3 \sqrt{5}}-3=\sqrt{61+24 \sqrt{5}}, \quad \frac{4-21+9 \sqrt{5}}{7-3 \sqrt{5}}=\sqrt{61+24 \sqrt{5}} \\
& \frac{9 \sqrt{5}-17}{7-3 \sqrt{5}}=\sqrt{61+24 \sqrt{5}}, \quad \frac{(9 \sqrt{5}-17)(7+3 \sqrt{5})}{(7-3 \sqrt{5})(7+3 \sqrt{5})}=\sqrt{61+24 \sqrt{5}} \\
& \frac{12 \sqrt{5}+16}{7^{2}-(3 \sqrt{5})^{2}}=\sqrt{61+24 \sqrt{5}}, \quad \frac{12 \sqrt{5}+16}{4}=\sqrt{61+24 \sqrt{5}} \\
& 3 \sqrt{5}+4=\sqrt{61+24 \sqrt{5}}
\end{aligned}
\]

Square both sides of the last equation. Then

\[
\begin{aligned}
& (3 \sqrt{5}+4)^{2}=61+24 \sqrt{5}, \quad 45+24 \sqrt{5}+16=61+24 \sqrt{5} \\
& 61+24 \sqrt{5}=61+24 \sqrt{5}
\end{aligned}
\]