Solve the following equation:

$$
\frac{1}{\sqrt{2+x}-\sqrt{2-x}}+\frac{1}{\sqrt{2+x}+\sqrt{2-x}}=1
$$
Multiplying the equation by the product of the two denominators, we obtain

$$
\sqrt{2+x}+\sqrt{2-x}+\sqrt{2+x}-\sqrt{2-x}=(2+x)-(2-x)
$$

which simplifies to

$$
2 \sqrt{2+x}=2 x
$$

and thus

$$
\sqrt{2+x}=x
$$

Squaring and rearranging,

$$
x^{2}-x-2=0
$$

from which

$$
x_{1}=2, \quad\left[x_{2}=-1\right]
$$

By substitution, we can verify that only the root $x_{1}=2$ satisfies the original equation. The root $x_{2}=-1$ is an "extraneous" root, which satisfies only the equation obtained by squaring but not the original square root equation. (If +1 on the right-hand side of the original equation were replaced by -1, then $x_{1}=-1$ would satisfy it, and $x_{2}=2$ would be the "extraneous" root.)

Kovács László (Debrecen, Ref. g. II. o. t.)