Đáp án+Giải thích các bước giải:
b,
`A=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{100}+\sqrt{99}}`
`A=\frac{\sqrt{2}-1}{(\sqrt{2}-1)(\sqrt{2}+1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})}+...+\frac{\sqrt{100}-\sqrt{99}}{(\sqrt{100}-\sqrt{99})(\sqrt{100}+\sqrt{99})}`
`A=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{2-1}+\frac{\sqrt{4}-\sqrt{3}}{2-1}+...+\frac{\sqrt{100}-\sqrt{99}}{2-1}`
`A=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}+...+\frac{\sqrt{100}-\sqrt{99}}{1}`
`A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}`
`A=-1+\sqrt{100}`
`A=-1+10`
`A=9`
