7. Given real numbers $a>b>0$. Then the maximum value of the function
$$
f(x)=\frac{x}{\sqrt{a-x^{2}}-\sqrt{b-x^{2}}}
$$

is . $\qquad$
7. $\frac{\sqrt{a b}}{a-b}$.

Obviously, when $x=0$, $f(x)>0$.
Therefore, we only need to consider the case when $x>0$.
Notice that,
$$
\begin{array}{l}
f(x)=\frac{x}{\sqrt{a-x^{2}}-\sqrt{b-x^{2}}} \\
=\frac{x\left(\sqrt{a-x^{2}}+\sqrt{b-x^{2}}\right)}{a-b} \\
=\frac{1}{a-b} \sqrt{x^{2}(a+b)-2 x^{4}+2 x^{2} \sqrt{\left(a-x^{2}\right)\left(b-x^{2}\right)}} \\
=\frac{1}{a-b} \sqrt{a b-\left(x^{2}-\sqrt{\left(a-x^{2}\right)\left(b-x^{2}\right)}\right)^{2}} \\
\leqslant \frac{\sqrt{a b}}{a-b} .
\end{array}
$$

Thus, when $x^{2}=\sqrt{\left(a-x^{2}\right)\left(b-x^{2}\right)}$, i.e., $x=\sqrt{\frac{a b}{a+b}}$,
the equality holds.