4. Given three real

numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is.
4.5

From the problem, we have
$$
\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{1} x_{3}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.

Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or $x_{2}=\frac{1}{5}$.

Therefore, $\left(x_{1}, x_{2}, x_{3}\right)$ has 5 sets.