Let A be the sum,then :

\(-A=\dfrac{-1}{1+\sqrt{2}}+\dfrac{-1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{-1}{\sqrt{98}+\sqrt{99}}+\dfrac{-1}{\sqrt{99}+\sqrt{100}}\)

With \(n\ge0\),we have :

\(\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=\left(n+1\right)-n=1\)

\(\Rightarrow\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)

\(\Rightarrow\dfrac{-1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n}-\sqrt{n+1}\)

Hence,we have :

\(-A=1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{98}-\sqrt{99}+\sqrt{99}-\sqrt{100}\)

\(=1-10=-9\)

\(\Rightarrow A=9\)