1. Real numbers $x, y, z$ (non-zero) satisfy the equation: $x+y=z$. Find all possible values that the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}$ can take.
1. Answer: 3.

Solution. Transform the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}=$

$=\frac{z}{x}-\frac{y}{x}+\frac{z}{y}-\frac{x}{y}+\frac{y}{z}+\frac{x}{z}=\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}$.

Using the condition $x+y=z$ and the derived formulas $y=z-x$ and $x=$ $z-y$, we get:

$\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}=\frac{x}{x}+\frac{y}{y}+\frac{z}{z}=3$.

Grading criteria:

Correct answer with justification provided: 7 points.

Only the correct answer provided without justification: 0 points.