Let's rationalize the denominator of the following fraction:

$$
\frac{1}{1+\sqrt[4]{32}-5 \sqrt{2}}
$$
I. solution. Since $\sqrt[4]{32}=2 \sqrt[4]{2}$, we first try to transform this so that only $\sqrt{2}$ terms remain:

$$
\frac{1}{1+\sqrt[4]{32}-5 \sqrt{2}}=\frac{1-5 \sqrt{2}-\sqrt[4]{32}}{(1-5 \sqrt{2})^{2}-\sqrt{32}}=\frac{1-5 \sqrt{2}-\sqrt[4]{32}}{51-14 \sqrt{2}}
$$

Now we can easily rationalize the denominator by multiplying the fraction by $51+14 \sqrt{2}$. Thus,

$$
\frac{1}{1+\sqrt[4]{32}-5 \sqrt{2}}=\frac{(1-5 \sqrt{2}-\sqrt[4]{32})(51+14 \sqrt{2})}{51^{2}-2 \cdot 196}=\frac{(51+14 \sqrt{2})(1-5 \sqrt{2}-\sqrt[4]{32})}{2209}
$$

Majthényi Ágnes (Győr, Révai M. Gymn, III o. t)

II. solution. The denominator of the fraction is of the form $a+b \sqrt[4]{2}+c(\sqrt[4]{2})^{2}+d(\sqrt[4]{2})^{3}$, since $1+\sqrt[4]{32}-5 \sqrt{2}=1+2 \sqrt[4]{2}-5(\sqrt[4]{2})^{2}$. Imagine that we have successfully rationalized the fraction. We will always get a number of this form. Therefore, we need to find rational numbers $a, b, c, d$ such that

$$
\left(a+b \sqrt[4]{2}+c(\sqrt[4]{2})^{2}+d(\sqrt[4]{2})^{3}\right)\left(1+2 \sqrt[4]{2}-5(\sqrt[4]{2})^{2}\right)=1
$$

After multiplying everything out, the coefficient of $\sqrt[4]{2}$ is $b+2 a-10 d$, the coefficient of $(\sqrt[4]{2})^{2}=\sqrt{2}$ is $2 b+c-5 a$, the coefficient of $(\sqrt[4]{2})^{3}$ is $d-2 c-5 b$, and the "rational part" is $a-10 c-4 d$. If we can find rational numbers $a, b, c, d$ such that

$$
\begin{aligned}
& 2 a+b-10 d=0 \\
& -5 a+2 b+c=0 \\
& -5 b+2 c+d=0 \\
& a-10 c+4 d=1
\end{aligned}
$$

then we have solved the problem. Solving the system of equations using any known method, we get $a=\frac{-89}{2209}, b=\frac{-102}{2209}, c=\frac{-241}{2209}, d=\frac{-28}{2209}$ as the solution. Therefore, $\frac{1}{1+\sqrt[4]{32}-5 \sqrt{2}}=-\frac{89+102 \sqrt[4]{2}+241 \sqrt{2}+28 \sqrt[4]{8}}{2209}$.

Czikó András (Kisvárda, Bessenyei Gy. Gymn, IV o. t.)