1. Find $x$ and $y$ that satisfy the following equation:

$$
(x-y)^{2}+(y-2 \sqrt{x}+2)^{2}=\frac{1}{2}
$$
Answer: $x=1, y=\frac{1}{2}$.

## Solution:

Let $u=(\sqrt{x}-1)^{2}+1=x-2 \sqrt{x}+2 \geq 1, v=y-2 \sqrt{x}+2$.

Then

$$
y-x=v-u \quad \text { and the original equation becomes }
$$

$$
\begin{gathered}
(v-u)^{2}+v^{2}=\frac{1}{2} \\
v^{2}-2 u v+u^{2}-\frac{1}{2}=0
\end{gathered}
$$

Considering the last equation as a quadratic equation in terms of $v$, and taking into account the non-negativity of the discriminant, we get

$$
u^{2} \leq 1
$$

Therefore, $u=1, v=\frac{1}{2}$. Then $x=1, y=\frac{1}{2}$.