4. Let the increasing function $f(x)$ defined on $(0,+\infty)$ satisfy: for any $x \in(0,+\infty)$, we have $f(x)>-\frac{6}{x}$, and $f\left(f(x)+\frac{6}{x}\right)=5$.
Then $f(10)=$ $\qquad$
4. $\frac{12}{5}$.

Let $f(a)=5$ where $a>0$.
By the definition of an increasing function, we have
$$
\begin{array}{l}
f(a)+\frac{6}{a}=a \Rightarrow 5+\frac{6}{a}=a \Rightarrow a=3 \\
\Rightarrow f(x)+\frac{6}{x}=3 \Rightarrow f(10)=3-\frac{6}{10}=\frac{12}{5} .
\end{array}
$$