Let's simplify the following expression:

$$
\frac{a+x}{x(x-y)(x-z)}+\frac{a+y}{y(y-z)(y-x)}+\frac{a+z}{z(z-x)(z-y)}
$$
In the addition of the first two fractions, by taking the product $x y(x-y) \cdot(x-z)(y-z)$ as the common denominator, the sum of the numerators can be transformed as follows:

$$
\begin{gathered}
(a+x) y(y-z)-(a+y) x(x-z)=a\left(y^{2}-x^{2}\right)+ \\
+x y(y-x)-a z(y-x)=(y-x)(a x+a y-a z+x y)
\end{gathered}
$$

therefore, the sum of the first two fractions is:

$$
S=\frac{a z+a y-a z+x y}{x y(z-x)(y-z)}
$$

(We simplified by $y-x$, which is permissible because if it were 0, the given expression would be meaningless.) In the addition of $S$ and the third fraction, by taking $x y z(z-x)(y-z)$ as the common denominator, the sum of the numerators can be transformed as follows:

$$
(a x+a y-a z+x y) z-(a+z) x y=a y(z-x)-a z(z-x)=a(z-x)(y-z)
$$

therefore, taking advantage of the simplification opportunity,

$$
K=\frac{a}{x y z}
$$

provided that $x, y, z$ are distinct and non-zero numbers.

Tamás Banyó (Veszprém, Lovassy L. Gymnasium II. grade)

Note. The above calculation can be somewhat simpler if we notice that the second terms of the numerators can be omitted. Indeed, by breaking each fraction into the sum of two fractions, we can always simplify with the second term of the numerator, and the sum of these three terms disappears:

$$
\frac{1}{(x-y)(x-z)}+\frac{1}{(y-z)(y-x)}+\frac{1}{(z-x)(z-y)}=\frac{(y-z)-(x-z)+(x-y)}{(x-y)(x-z)(y-z)}=0
$$

László Polányi (Székesfehérvár, József A. Gymnasium, II. grade)