`P=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}`

`=\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}+\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+...+\frac{\sqrt{99}-\sqrt{100}}{(\sqrt{99}+\sqrt{100})(\sqrt{99}-\sqrt{100})}`

`=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+....+\frac{\sqrt{99}-\sqrt{100}}{99-100}`

`=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}`

`=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-.....-\sqrt{99}+\sqrt{99}-\sqrt{100}}{-1}`

`=\frac{\sqrt{1}-\sqrt{100}}{-1}`

`=\frac{\sqrt{1^2}-\sqrt{10^2}}{-1}`

`=\frac{1-10}{-1}`

`=\frac{-9}{-1}`

`=9`

Cách giải: Nhân lượng liên hợp `->` Rút gọn `->` Kết quả.

Đáp án:

Giải thích các bước giải:

$P=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\\=\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}+\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+...+\frac{\sqrt{99}-\sqrt{100}}{(\sqrt{99}+\sqrt{100})(\sqrt{99}-\sqrt{100})}\\=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\\=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-.....-\sqrt{99}+\sqrt{99}-\sqrt{100}}{-1}\\=\frac{\sqrt{1}-\sqrt{100}}{-1} =\frac{\sqrt{1^2}-\sqrt{10^2}}{-1} =\frac{1-10}{-1} = 9$