Solve the following equation in the set of real numbers:

$$
x^{2}+5 y^{2}+5 z^{2}-4 x z-2 y-4 y z+1=0 .
$$
We can transform the left side of the equation into the sum of complete squares as follows:

$$
(x-2 z)^{2}+(2 y-z)^{2}+(y-1)^{2}=0
$$

Since the square of any real number is a non-negative number, the sum on the left side can only be zero if each of its terms is zero. That is,

$$
(y-1)^{2}=0, \quad(2 y-z)^{2}=0, \quad(x-2 z)^{2}=0
$$

from which we get

$$
y=1, \quad z=2 y, \quad x=2 z
$$

that is,

$$
y=1, \quad z=2, \quad x=4
$$

Since we performed equivalent transformations throughout, the numbers obtained are indeed the roots of the equation.