已知实数 $a>b>0$，函数 $f(x)=\dfrac{x}{\sqrt{a-x^2}-\sqrt{b-x^2}}$ 的最大值是_______．
答案 $\dfrac{\sqrt{ab}}{a-b}$．
解析 根据题意，有\[\begin{split} f(x)&=\dfrac{x}{\sqrt{a-x^2}-\sqrt{b-x^2}}\\ &=\dfrac{x\left(\sqrt{a-x^2}+\sqrt{b-x^2}\right)}{a-b}\\ &\leqslant\dfrac{\sqrt{x^2(a-x^2)+x^2(b-x^2)+2x^2\sqrt{(a-x^2)(b-x^2)}}}{a-b}\\ &=\dfrac{\sqrt{ab-\left(x^2-\sqrt{(a-x^2)(b-x^2)}\right)^2}}{a-b}\\ &\leqslant \dfrac{\sqrt{ab}}{a-b},\end{split}\]等号当\[\begin{cases} x\geqslant 0,\\ x^2=\sqrt{(a-x^2)(b-x^2)},\end{cases} iff x=\sqrt{\dfrac{ab}{a+b}}\]时取得，因此所求最大值为 $\dfrac{\sqrt{ab}}{a-b}$．