11.7. Solve the equation $\sqrt{y \sqrt{5}}-\sqrt{x \sqrt{5}}=\sqrt{3 \sqrt{5}-5}$ in rational numbers.
Solution. We determine the DVA of this equation:

$$
\left\{\begin{array}{c}
\sqrt{y \sqrt{5}}-\sqrt{x \sqrt{5}} \geq 0 \\
y \geq 0, \quad x \geq 0
\end{array} \Leftrightarrow 0 \leq x \leq y\right.
$$

Let $x, y \in Q$. We square both sides of the equation:

$$
\begin{aligned}
& y \sqrt{5}-2 \sqrt{y \sqrt{5}} \cdot \sqrt{x \sqrt{5}}+x \sqrt{5}=3 \sqrt{5}-5 \\
& (y+x-3) \cdot \sqrt{5}=2 \sqrt{5 x y}-5 \\
& y+x-3=2 \sqrt{x y}-\sqrt{5}
\end{aligned}
$$

We square again:

$$
(y+x-3)^{2}=4 x y-4 \sqrt{x y} \cdot \sqrt{5}+5 \Rightarrow 4 \sqrt{5 x y}=\left(4 x y+5-(y+x-3)^{2}\right) \in Q
$$

It follows that $\sqrt{5 x y} \in Q$. From $(*)$ we get that $2 \sqrt{5 x y}-5 \in Q$. Since on the left side of equation $(*)$ for $y+x-3 \neq 0$ we have an irrational number, then the system of equations must hold

$$
\left\{\begin{array}{c}
y+x-3=0 \\
2 \sqrt{5 x y}-5=0
\end{array}\right.
$$

We solve this system and obtain 2 solutions: $(x ; y)=\left(\frac{5}{2} ; \frac{1}{2}\right)$ or $(x ; y)=\left(\frac{1}{2} ; \frac{5}{2}\right)$. Considering DVA, only the solution $(x ; y)=\left(\frac{1}{2} ; \frac{5}{2}\right)$ remains.