Let's find all the real number pairs $(x, y)$ for which the following holds:

$$
\sqrt{\frac{1}{4-x^{2}}}+\sqrt{\frac{y^{2}}{y-1}}=\frac{5}{2}
$$
The first term on the left side is only defined for $|x|1$. The expression under the square root sign is the reciprocal of $1 / y$ and the product of $1-1 / y$, where $1 / y$ is a positive number less than one. The arithmetic mean of these two quantities is $1 / 2$, so their geometric mean is equal to or less than $1 / 2$. The value of our expression is the reciprocal of the geometric mean, so its value is either equal to 2 or greater than 2.

If, therefore, $y>1$, then the value of the first term on the left side of (1) is at least $1 / 2$, and the second is at least 2. The sum of the two terms can only be $5 / 2$ if the first is exactly $1 / 2$ and the second is exactly 2, that is, $x=0, y=2$. For the variable $y$, apart from the values $y>1$, only $y=0$ can come into question, for which we need to choose an $x$ such that $4-x^{2}=4 / 25, x^{2}=96 / 25$, $x= \pm \sqrt{96 / 5}$. Thus, (1) is satisfied by three pairs of numbers: $x=0, y=2 ; x=\sqrt{96,5}, y=0$ and $x=-\sqrt{96,5}, y=0$.