Let's determine the real roots of the following equation:

$$
\frac{1+x^{4}}{(1+x)^{4}}=\frac{1}{2}
$$
I. Solution: Multiply both sides of the equation by $2(1+x)^{4}$ (for $x \neq-1$) and transform the right side into a polynomial:

$$
2+2 x^{4}=1+4 x+6 x^{2}+4 x^{3}+x^{4}
$$

that is,

$$
x^{4}-4 x^{3}-6 x^{2}-4 x+1=0
$$

Divide the equation by $x^{2}$ (for $x \neq 0$) and regroup the terms:

$$
x^{2}+\frac{1}{x^{2}}-4 x-\frac{4}{x}-6=0
$$

or

$$
\left(x^{2}+\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)-6=0
$$

Let $x+\frac{1}{x}=y$, then $y^{2}=x^{2}+\frac{1}{x^{2}}+2$,

and thus

$$
y^{2}-2-4 y-6=0
$$

or

$$
y^{2}-4 y-8=0
$$

from which

$$
y_{1}=2+2 \sqrt{3}, \quad y_{2}=2-2 \sqrt{3}
$$

In the first case,

$$
x+\frac{1}{x}=2+2 \sqrt{3}
$$

or

$$
x^{2}-(2+2 \sqrt{3}) x+1=0
$$

from which

$$
x_{1}=1+\sqrt{3}+\sqrt{3+2 \sqrt{3}}, \quad x_{2}=1+\sqrt{3}-\sqrt{3+2 \sqrt{3}}
$$

In the second case,

$$
x+\frac{1}{x}=2-2 \sqrt{3}
$$

or

$$
x^{2}-(2-2 \sqrt{3}) x+1=0
$$

from which

$$
x_{3}=1-\sqrt{3}+\sqrt{3-2 \sqrt{3}}, \quad x_{4}=1-\sqrt{3}-\sqrt{3-2 \sqrt{3}}
$$

Clearly, $x_{1}$ and $x_{2}$ are real roots, while $x_{3}$ and $x_{4}$ are complex roots, because $3-2 \sqrt{3}<0$, and thus $\sqrt{3-2 \sqrt{3}}$ is imaginary. For positive $x$, $x+\frac{1}{x} \geq 2$, and for negative $x$, $x+\frac{1}{x} \leq -2$. Therefore, $x_{3}$ and $x_{4}$ do not satisfy the original equation.