$$
\begin{array}{l}
\log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\
=2 \log _{3}(5 x+2) \log _{5}(5 x+4)
\end{array}
$$

Solve the equation for \( x \):

$$
\begin{array}{l}
\log _{2}(5 x+1) \log _{5}(5 x+4)+\log _{3}(5 x+2) \log _{4}(5 x+3) \\
=2 \log _{3}(5 x+2) \log _{5}(5 x+4)
\end{array}
$$
First, we prove the lemma: If \( n \geqslant m > 1 \), then
\[
01, \\
\therefore \log _{n}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{n}\right) \leqslant \log _{m}\left(1+\frac{1}{m}\right), \\
\therefore \log _{n}(n+1)-1 \leqslant \log _{m}(m+1)-1, \\
\log _{n}(n+1) \leqslant \log _{m}(m+1) . \\
\frac{\lg (n+1)}{\lg n} \leqslant \frac{\lg (m+1)}{\lg m}, \\
\frac{\lg m}{\lg n} \leqslant \frac{\lg (m+1)}{\lg (n+1)},
\end{array}
\]
Thus,
\[
\log _{n} m \leqslant \log _{(n+1)}(m+1).
\]
Second, it is easy to verify that \( x = \frac{1}{5} \) is a root of the original equation.
Finally, we prove that the original equation has no other real roots:
Let \( f(x) = \frac{\log _{2}(5 x+1)}{\log _{3}(5 x+2)} + \frac{\log _{4}(5 x+3)}{\log _{5}(5 x+4)} \). Then the original equation can be transformed into \( f(x) = 2 \). Since \( 5 x + 1 > 0 \), it follows that \( 5 x + 2, 5 x + 3, 5 x + 4 \) are all greater than 1. According to the lemma:
If \( 0 < 5 x + 1 < 2 \), then
\[
f(x) = \frac{\log _{7}(5 x+2)}{\log _{2}(5 x+1)} + \frac{\log _{5}(5 x+4)}{\log _{4}(5 x+3)} > 1 + 1 = 2.
\]
If \( 5 x + 1 > 2 \), then
\[
f(x) = \frac{\log _{7}(5 x+2)}{\log _{2}(5 x+1)} + \frac{\log _{5}(5 x+4)}{\log _{4}(5 x+3)} < 1 + 1 = 2.
\]
In summary, when \( 0 < 5 x + 1 \neq 2 \), \( f(x) \neq 2 \).
Therefore, the original equation has and only has one real root \( x = \frac{1}{5} \).