Solve the following equations:
a) $2+\frac{5}{4 x}-\frac{15}{4 x(8 x+3)}=\frac{2(7 x+1)}{7 x-3}$,
b) $\frac{2}{x}+\frac{1}{x^{2}}-\frac{7+10 x}{x^{2}\left(x^{2}+7\right)}=\frac{2}{x+\frac{3}{x+\frac{4}{x}}}$.
a) Assuming that none of $x, 8 x+3, 7 x-3$ are zero, multiplying both sides of the equation by $4 x(8 x+3)(7 x-3)$

$$
8 x(8 x+3)(7 x-3)+5(8 x+3)(7 x-3)-15(7 x-3)=8 x(7 x+1)(8 x+3)
$$

Transforming into a polynomial and rearranging

$$
24 x^{2}-216 x=0, \quad \text { or } \quad x^{2}-9 x=0
$$

Since the value $x=0$ is excluded, the only root is

$$
x=9
$$

b) The right side, assuming that $x \neq 0$ and $x^{2}+4 \neq 0$, can be transformed as follows:

$$
\frac{2}{x+\frac{3 x}{x^{2}+4}}=\frac{2}{\frac{x^{3}+7 x}{x^{2}+4}}=\frac{2 x^{2}+8}{x\left(x^{2}+7\right)}
$$

Thus, our equation becomes

$$
\frac{2}{x}+\frac{1}{x^{2}}-\frac{7+10 x}{x^{2}\left(x^{2}+7\right)}=\frac{2 x^{2}+8}{x\left(x^{2}+7\right)}
$$

Assuming further that $x^{2}+7 \neq 0$, the fractions can be eliminated:

$$
2 x^{3}+14 x+x^{2}+7-7-10 x=2 x^{3}+8 x
$$

Rearranging

$$
x^{2}-4 x=0
$$

Since the value $x=0$ is excluded, the only root is

$$
x=4
$$

Jármay Erzsébet (Bp. XIII., Ép. gép. techn. II. o. t.)