Question

Chứng minh:

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

Answer 1

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

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

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

Do đó : \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)