Solve the following equations:
a) $1+\frac{1}{2+\frac{1}{\frac{4 x+1}{2 x+1}-\frac{1}{2+\frac{1}{x}}}}=\frac{19}{14}$
b) $\frac{\frac{2 x-1}{2}+\frac{4}{3}}{\frac{x-1}{3}-\frac{1}{2}\left(1-\frac{1}{3}\right)}-\frac{x+4}{\frac{2 x+1}{2}+\frac{1}{5}-2-\frac{1}{1+\frac{1}{2+\frac{1}{3}}}}=\frac{9-2 x}{2 x-4}$.
a) The left side can be transformed as follows (if $x \neq 0, -\frac{1}{2}, -\frac{1}{3}, -\frac{3}{8}$):

$$
1+\frac{1}{2+\frac{1}{\frac{3 x+1}{2 x+1}}}=1+\frac{1}{\frac{8 x+3}{3 x+1}}=\frac{11 x+4}{8 x+3}
$$

Thus,

$$
\frac{11 x+4}{8 x+3}=\frac{19}{14}
$$

which means

$$
154 x+56=152 x+57
$$

from which

$$
x=\frac{1}{2} \text{.}
$$

b) Transforming the left side:

$$
\begin{gathered}
\frac{6 x+5}{2 x-4}-\frac{x+4}{\frac{2 x+1}{2}+\frac{1}{5}-2-\frac{7}{10}}=\frac{6 x+5}{2 x-4}-\frac{x+4}{\frac{10 x-20}{10}}= \\
=\frac{6 x+5}{2 x-4}-\frac{2 x+8}{2 x-4}=\frac{4 x+3}{2 x-4}
\end{gathered}
$$

Thus,

$$
\frac{4 x+3}{2 x-4}=\frac{9-2 x}{2 x-4}
$$

Assuming $2 x-4 \neq 0$; that is, $x \neq 2$, it is safe to multiply both sides of the equality by $(2 x-4)$:

$$
4 x+3=9-2 x
$$

from which

$$
x=2,
$$

which is precisely the excluded value.

Thus, we have a contradiction, as there is no value of $x$ that satisfies our equation.

Szabó Endre (Gyöngyös, Vak Bottyán g. II. o. t.)

Note: The Corvin Mátyás g. specialized group graphically demonstrated that there is no root by plotting both sides of the equality under (1) as functions of $x$. (The line $x=2$ is a common asymptote for the two hyperbolas.)