1. For what values of $x$ and $y$ is the equality $x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3}$ true?
Solution. Expanding the brackets and grouping the terms by powers of the variable $x$, we obtain a quadratic equation in $x$ of the following form: $x^{2}-y x+\left(y^{2}-y+\frac{1}{3}\right)^{2}=0$. Then $D=y^{2}-4 y^{2}+4 y-\frac{4}{3}=-3 y^{2}+4 y-\frac{4}{3}=-3\left(y-\frac{2}{3}\right)^{2}$. The quadratic equation has roots if $D \geq 0$, therefore, $y=\frac{2}{3}$. Substituting this value into the original equation, we get that $x=\frac{1}{3}$.

Answer: $x=\frac{1}{3}, y=\frac{2}{3}$.