2.222. $\left(\frac{2-n}{n-1}+4 \cdot \frac{m-1}{m-2}\right):\left(n^{2} \cdot \frac{m-1}{n-1}+m^{2} \cdot \frac{2-n}{m-2}\right) ; m=\sqrt[4]{400}, n=\sqrt{5}$.
Solution.

$$
\begin{aligned}
& \left(\frac{2-n}{n-1}+4 \cdot \frac{m-1}{m-2}\right):\left(n^{2} \cdot \frac{m-1}{n-1}+m^{2} \cdot \frac{2-n}{m-2}\right)= \\
& =\frac{(2-n)(m-2)+4(m-1)(n-1)}{(n-1)(m-2)}: \frac{n^{2}(m-1)(m-2)+m^{2}(2-n)(n-1)}{(n-1)(m-2)}= \\
& =\frac{3 m n-2(m+n)}{(n-1)(m-2)}: \frac{(m-n)(3 m n-2(m+n))}{(n-1)(m-2)}= \\
& =\frac{3 m n-2(m+n)}{(n-1)(m-2)} \cdot \frac{(n-1)(m-2)}{(m-n)(3 m n-2(m+n))}=\frac{1}{m-n}=\frac{1}{\sqrt[4]{400}-\sqrt{5}}= \\
& =\frac{1}{\sqrt{20}-\sqrt{5}}=\frac{1}{2 \sqrt{5}-\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}
\end{aligned}
$$

Answer: $\frac{\sqrt{5}}{5}$.