Question

Bài 1: CM: \(\frac{1}{(n+1)+\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n+1}}\)

Bài 2: áp dụng bài 1 tính:

\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

Answer 1

(Fix luôn lại đề)

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

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

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

Bài 2:

Áp dụng bài 1 vào A được:

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