Solve the following equation:

$$
\frac{\frac{5-4 x}{5+4 x}+3}{3+\frac{5+4 x}{5-4 x}}-\frac{\frac{5-4 x}{5+4 x}+2}{2+\frac{5+4 x}{5-4 x}}=\frac{\frac{5-4 x}{5+4 x}+1}{1+\frac{5+4 x}{5-4 x}}
$$
I. solution. It is natural to introduce a new unknown for the quotient appearing in each of the (large) numerators. Let

$$
\begin{aligned}
\frac{5-4 x}{5+4 x}= & y, \quad \text { so } \quad \frac{5+4 x}{5-4 x}=\frac{1}{y}, \quad \text { and } \\
& \frac{y+3}{3+\frac{1}{y}}-\frac{y+2}{2+\frac{1}{y}}=\frac{y+1}{1+\frac{1}{y}}
\end{aligned}
$$

from which, by the usual rearrangement steps,

$$
7 y^{2}+5 y=y(7 y+5)=0
$$

In the process, we assumed that the multipliers used to eliminate the fractions: $y, 3 y+1, 2 y+1, y+1$ are not 0, i.e., $y \neq 0$, $-1 / 3, -1 / 2, -1$.

Accordingly, (2) can only be satisfied if

$$
7 y+5=0, \quad y=-5 / 7
$$

which is not an excluded value. Now,

$$
\frac{5-4 x}{5+4 x}=-\frac{5}{7} \text {, so } \quad x=15 / 2=7.5
$$

With this, the value of both sides of (1) is $-5 / 7$, so the solution is correct.

László Márki (Budapest, Fazekas M. Gyak. G. I. o. t.)

II. solution. We can solve our equation without introducing a new unknown. If both sides of (1) make sense, we can multiply them by $(5+4 x) /(5-4 x)$:

$$
\frac{20+8 x}{20-8 x}-\frac{15+4 x}{15-4 x}=\frac{10}{10}=1
$$

which, when multiplied by $(5-2 x)(15-4 x)$ and rearranged, gives

$$
8 x^{2}-70 x+75=0, \quad x_{1}=5 / 4, \quad x_{2}=15 / 2
$$

$x_{1}$ makes the common second term of the (large) denominators in (1) meaningless, so only $x_{2}$ can be a root. This, as we have seen, is indeed a solution.

Péter Gangli (Apáczai Csere J. Gyak. G. II. o. t.)