7. If real numbers $x, y$ satisfy $x-4 \sqrt{y}=2 \sqrt{x-y}$, then the range of values for $x$ is $\qquad$ .
7. $\{0\} \cup[4,20]$.

Let $\sqrt{y}=a, \sqrt{x-y}=b(a, b \geqslant 0)$, at this point,
$$
x=y+(x-y)=a^{2}+b^{2},
$$

and the given equation becomes
$$
a^{2}+b^{2}-4 a=2 b.
$$
Thus, $a, b$ satisfy the equation
$$
(a-2)^{2}+(b-1)^{2}=5(a, b \geqslant 0).
$$

As shown in Figure 3, in the $aOb$ plane, the trajectory of point $(a, b)$ is the part of the circle with center $D(1,2)$ and radius $\sqrt{5}$ that lies in the region $a, b \geqslant 0$, i.e., the union of point $O$ and arc $\overparen{ACB}$.
$$
\begin{array}{c}
\text{Therefore, } \sqrt{a^{2}+b^{2}} \\
\in\{0\} \cup[2,2 \sqrt{5}].
\end{array}
$$

Thus, $x=a^{2}+b^{2} \in\{0\} \cup[4,20]$.