Let's calculate the value of the following expression:

$$
\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}} \quad \text { if } \quad x=\frac{\sqrt{3}}{2}
$$
First solution. Let

$$
\sqrt{1+x}=a \quad \text { and } \quad \sqrt{1-x}=b
$$

then

$$
(a-b)^{2}=(\sqrt{1+x}-\sqrt{1-x})^{2}=2-2 \sqrt{1-x^{2}}=2-2 \sqrt{1-\frac{3}{4}}=1
$$

thus

$$
a-b=1
$$

Furthermore,

$$
a^{2}+b^{2}=(1+x)+(1-x)=2
$$

and

$$
a b=\frac{a^{2}+b^{2}-(a-b)^{2}}{2}=\frac{1}{2}
$$

Therefore,

$$
\begin{gathered}
\frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=\frac{a^{2}}{1+a}+\frac{b^{2}}{1-b}= \\
=\frac{a^{2}+b^{2}-a b(a-b)}{1+(a-b)-a b}=\frac{2-\frac{1}{2}}{1+1-\frac{1}{2}}=1
\end{gathered}
$$

Second solution. If we rationalize the denominators, then multiply the numerator and denominator by $x$, and finally substitute the value of $x$, we get:

$$
\frac{1}{3}\left[-6+2 \sqrt{3}\left(\sqrt{1+\sqrt{\frac{3}{4}}}+\sqrt{1-\sqrt{\frac{3}{4}}}\right)+3\left(\sqrt{1+\sqrt{\frac{3}{4}}}-\sqrt{1-\sqrt{\frac{3}{4}}}\right)\right]
$$

But

$$
\sqrt{1+\sqrt{\frac{3}{4}}}+\sqrt{1-\sqrt{\frac{3}{4}}}=\sqrt{\left(\sqrt{1+\sqrt{\frac{3}{4}}}+\sqrt{1-\sqrt{\frac{3}{4}}}\right)^{2}}=\sqrt{3}
$$

and

$$
\sqrt{1+\sqrt{\frac{3}{4}}}-\sqrt{1-\sqrt{\frac{3}{4}}}=\sqrt{\left(\sqrt{1+\sqrt{\frac{3}{4}}}-\sqrt{1-\sqrt{\frac{3}{4}}}\right)^{2}}=1
$$

Substituting these values, we again get the result of 1.

(Basch Rudolf, Budapest.)

The problem was also solved by: Erdős V., Fried E., Koffler B., Szántó L.