jmc

Let $\alpha =\sqrt[4]{n+1}$ and $\beta =\sqrt[4]{n}.$ Then $$\begin{array}{rl}\frac{1}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{n}+\sqrt[4]{n+1})}& =\frac{1}{({\alpha }^2+{\beta }^2)(\alpha +\beta )}\\ & =\frac{\alpha -\beta }{({\alpha }^2+{\beta }^2)(\alpha +\beta )(\alpha -\beta )}\\ & =\frac{\alpha -\beta }{({\alpha }^2+{\beta }^2)({\alpha }^2-{\beta }^2)}\\ & =\frac{\alpha -\beta }{{\alpha }^4-{\beta }^4}\\ & =\frac{\alpha -\beta }{(n+1)-n}\\ & =\alpha -\beta \\ & =\sqrt[4]{n+1}-\sqrt[4]{n}.\end{array}$$Hence, $$\begin{array}{rl}\sum _{n=1}^{9999}\frac{1}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{n}+\sqrt[4]{n+1})}& =(\sqrt[4]{2}-\sqrt[4]{1})+(\sqrt[4]{3}-\sqrt[4]{2})+\cdots +(\sqrt[4]{10000}-\sqrt[4]{9999})\\ & =\sqrt[4]{10000}-\sqrt[4]{1}=\boxed{9}.\end{array}$$