Let's simplify the following fraction:

$$
\frac{x y^{2}+2 y z^{2}+y z u+2 x y z+2 x z u+y^{2} z+2 z^{2} u+x y u}{x u^{2}+y z^{2}+y z u+x u z+x y u+u z^{2}+z u^{2}+x y z}
$$
Both the numerator and the denominator can be paired in such a way that by factoring out the common letters, $(x+z)$ remains in the parentheses, and the fraction thus becomes:

$$
K=\frac{y^{2}(x+z)+2 y z(z+x)+y u(z+x)+2 z u(x+z)}{u^{2}(x+z)+y z(z+x)+y u(z+x)+u z(x+z)}
$$

After factoring out the common factor $(x+z)$ from the resulting 4-term expressions, similar pairing can be done, and $y+u$ can be factored out.

$$
K=\frac{(x+z)\left[\left(y^{2}+y u\right)+(2 y z+2 z u)\right]}{(x+z)\left[\left(u^{2}+y u\right)+(y z+u z)\right]}=\frac{(x+z)(y+u)(y+2 z)}{(x+z)(u+y)(u+z)}
$$

Therefore, $K=(y+2 z) /(u+z)$ for every value system of $x, y, z, u$ where the product of the common factors of the original numerator and denominator is not zero, i.e., $x+z \neq 0, y+u \neq 0$, or in other words, $z \neq -x, u \neq -y$. If these two conditions are not met, then the original form of $k$ is of the form $0 / 0$, and thus is undefined.

János Mogyorósi (Kóvágóörs, General School VIII. grade)