1. Find all pairs of real numbers (x, y) for which the equality $\sqrt{x^{2}+y^{2}-1}=x+y-1$ holds.
Answer: All pairs of numbers $(1, t),(t, 1)$, where $t$ is any non-negative number.

Hint. By squaring and factoring, we get that $\mathrm{x}=1$ or $\mathrm{y}=1$. The square root is non-negative, so the sum $x+y-1 \geq 0$. Substituting the possible values of the variables into this inequality, we get that the other variable is non-negative. Criteria. Not taken into account that the arithmetic root is non-negative: 3 points.

One set of pairs is lost: 1 point.