7. 3 .

Analysis: Since $f(x)=\frac{5 x+\sqrt{9 x^{2}-16}}{1+x^{2}}=\frac{16}{5 x-\sqrt{9 x^{2}-16}}$, we only need to consider the range of $g(x)=5 x-\sqrt{9 x^{2}-16}$. $g^{\prime}(x)=5-\frac{9 x}{\sqrt{9 x^{2}-16}} \geq 0 \Leftrightarrow x \geq \frac{5}{3}$, so $g(x) \geq g\left(\frac{5}{3}\right)=\frac{16}{3}$, then $f(x) \leq \frac{16}{g(x)}=3$. When $x=\frac{5}{3}$, $f\left(\frac{5}{3}\right)=3$, hence the maximum value of the original function is 3.
7. The maximum value of the function $f(x)=\frac{5 x+\sqrt{9 x^{2}-16}}{1+x^{2}}$ is