`C=1/(\sqrt[]{1}+\sqrt[]{2})+1/(\sqrt[]{2}+\sqrt[]{3})+...+1/(\sqrt[]{99}+\sqrt[]{100})`
Nhân liên hiệp, ta có
`C=(\sqrt[]{2}-\sqrt[]{1})/[(\sqrt[]{2}-\sqrt[]{1})(\sqrt[]{2}+\sqrt[]{1})]+...+(\sqrt[]{100}-\sqrt[]{99})/[(\sqrt[]{100}-\sqrt[]{99})(\sqrt[]{100}+\sqrt[]{99})]`
`=\sqrt[]{2}-\sqrt[]{1}+\sqrt[]{3}-\sqrt[]{2}+...+\sqrt[]{99}-\sqrt[]{98}+\sqrt[]{100}-\sqrt[]{99}`
`=-\sqrt[]{1}+\sqrt[]{100}`
`=\sqrt[]{100}-\sqrt[]{1}`
`=10-1=9`
