Solve the following equation:

$$
\frac{\frac{2 x+1}{2 x-1}-1}{1-\frac{2 x-1}{2 x+1}}+\frac{\frac{2 x+1}{2 x-1}-2}{2-\frac{2 x-1}{2 x+1}}+\frac{\frac{2 x+1}{2 x-1}-3}{3-\frac{2 x-1}{2 x+1}}
$$
I. solution: The values of the denominators that are 0 must be excluded, because otherwise our equation would be meaningless. So, $2 x \pm 1 \neq 0$, that is, $x \neq \pm \frac{1}{2}$

$1-\frac{2 x-1}{2 x+1}=\frac{2}{2 x+1} \neq 0$ always holds; $2-\frac{2 x-1}{2 x+1}=\frac{2 x+3}{2 x+1} \neq 0$,

that is, $x \neq-\frac{3}{2}$, and finally $3-\frac{2 x-1}{2 x+1}=\frac{4 x+4}{2 x+1} \neq 0$, that is, $x \neq-1$.

Considering these excluded values, the terms of our equation can be transformed as follows:

The first term on the left side:

$$
\frac{\frac{2 x+1-(2 x-1)}{2 x-1}}{\frac{2 x+1-(2 x-1)}{2 x+1}}=\frac{\frac{2}{2 x-1}}{\frac{2}{2 x+1}}=\frac{2 x+1}{2 x-1}
$$

The second term on the left side:

$$
\frac{\frac{2 x+1-(4 x-2)}{2 x-1 x}}{\frac{4 x+2-(2 x-1)}{2 x+1}}=\frac{\frac{3-2 x}{2 x-1}}{\frac{2 x+3}{2 x+1}}=\frac{(2 x+1)(3-2 x)}{(2 x-1)(2 x+3)}
$$

The right side:

$$
\frac{\frac{2 x+1-(6 x-3)}{2 x-1}}{\frac{6 x+3-(2 x-1)}{2 x+1}}=\frac{\frac{4-4 x}{2 x-1}}{\frac{4 x+4}{2 x+1}}=\frac{(2 x-1)(4-4 x)}{(2 x-1)(4 x+4)}
$$

Thus, our equation takes the following form:

$$
\frac{2 x+1}{2 x-1}+\frac{2 x+1}{2 x-1} \cdot \frac{3-2 x}{2 x+3}=\frac{2 x+1}{2 x-1} \cdot \frac{4-4 x}{4 x+4}
$$

Considering that we have excluded the values $x= \pm \frac{1}{2}$, $\frac{2 x+1}{2 x-1}$ is a non-zero value, by which we can divide our equation, and simplifying the right side by 4, we get

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

Since we have already assumed that $x \neq-\frac{3}{2}$ and $x \neq-1$, eliminating the fractions,

$$
(x+1)(2 x+3)+(3-2 x)(x+1)=(1-x)(2 x+3)
$$

or

$$
2 x^{2}+7 x+3=0
$$

from which

$$
x_{1}=-3, \quad\left[x_{2}=-\frac{1}{2}\right]
$$

The latter root we have excluded, so the only root of our equation is

$$
x_{1}=-3
$$

Argay Gyula (Balassagyarmat, Balassi Bálint g. II. o. t.)

II. solution: Let $\frac{2 x-1}{2 x+1}=y$. Assuming that $x \neq \pm \frac{1}{2}$, $y$ has a meaning and is a non-zero value.

Performing this substitution, our equation becomes:

$$
\frac{\frac{1}{y}-1}{1-y}+\frac{\frac{1}{y}-2}{2-y}=\frac{\frac{1}{y}-3}{3-y}
$$

or

$$
\frac{1-y}{y} \cdot \frac{1}{1-y}+\frac{1-2 y}{y} \cdot \frac{1}{2-y}=\frac{1-3 y}{y} \cdot \frac{1}{3-y}
$$

Excluding the values $y=1, y=2$ and $y=3$, which correspond to the excluded values in the first solution, multiplying (1) by $y$,

$$
1+\frac{1-2 y}{2-y}=\frac{1-3 y}{3-y}
$$

Eliminating the fractions,

$$
(2-y)(3-y)+(1-2 y)(3-y)=(1-3 y)(2-y)
$$

or

$$
5 y=7,
$$

from which

$$
y=\frac{2 x-1}{2 x+1}=\frac{7}{5}
$$

From this,

$$
x=-3
$$

Könyves Tóth Pál (Bp. V., Eötvös g. II. o. t.)