The denominators of the following fractions need to be rationalized:

1. 

$$
\frac{a}{b-\sqrt{c}+a}
$$

2. 

$$
\frac{a}{x+\sqrt{y}+z+\sqrt{u}}
$$

3. 

$$
2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
$$
1. 

$$
\frac{a}{b-\sqrt{c}+a}=a \frac{b+a+\sqrt{c}}{(b+a)^{2}-c}
$$

2. 

$$
\begin{gathered}
\frac{a}{x+\sqrt{y}+z+\sqrt{u}}=a \frac{x+z-(\sqrt{y}+\sqrt{u})}{(x+z)^{2}-(\sqrt{y}+\sqrt{u})^{2}}=a \frac{x+z-(\sqrt{y}+\sqrt{u})}{(x+z)^{2}-y-u-2 \sqrt{y u}}= \\
=a \frac{(x+z-\sqrt{y}-\sqrt{u})\left[(x+z)^{2}-y-u+2 \sqrt{y u}\right]}{\left[(x+z)^{2}-y-u\right]^{2}-4 y u}
\end{gathered}
$$

3. 

$$
\begin{gathered}
2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}= \\
=2^{4} \sqrt{2} \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2}} \cdot \frac{\sqrt{2-\sqrt{2}+\sqrt{2}}}{\sqrt{2-\sqrt{2}}} \cdot \frac{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}}{\sqrt{2-\sqrt{2+\sqrt{2}}}}=2^{4} \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}
\end{gathered}
$$

(Kirchknopf Ervin, Budapest.

The problem was also solved by: Dénes M., Erdős V., Fried E., Füstös P., Kiss J., Kovács Gy., Neumann L., Paunz A., Sárközy P., Szilárd V., Tandlich E., Viola R., Wáhl V.