37.2. It is known that $a>b>0$ and $a^{2}+b^{2}=6 a b$. Find $\frac{a+b}{a-b}$.

$$
(8-10 \text { th grade) }
$$
37.2. Given that $a>b>0$ and $a^{2}+b^{2}=6 a b$, we get:

$$
\frac{a+b}{a-b}=\sqrt{\left(\frac{a+b}{a-b}\right)^{2}}=\sqrt{\frac{a^{2}+2 a b+b^{2}}{a^{2}-2 a b+b^{2}}}=\sqrt{\frac{8 a b}{4 a b}}=\sqrt{2}
$$