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

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

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

Áp dụng vào A ta được

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

    \(=1-\frac{1}{10}\)

   \(=\frac{9}{10}\)

Incursion_03 đúng mẹ nó rồi nhé!

tui cx định tl nhưng nó tl trước ns chung nó đúng cmnr

b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

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

Áp dụng vào bài toán ta được

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

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

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Cả 2 câu là n tự nhiên khác 0 hết nhé

a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)

Áp đụng vào bài toán được

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)

\(=\sqrt{1681}-\sqrt{1}=41-1=40\)

Chứng minh phụ: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) (trục căn thức ở mẫu)

                                   \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n^2+2n+1-n^2-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào tính: \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(\frac{1}{\left(1+1\right)\sqrt{1}+1\sqrt{1+1}}+\frac{1}{\left(1+2\right)\sqrt{2}+2\sqrt{2+1}}+...+\frac{1}{\left(99+1\right)\sqrt{99}+99\sqrt{99+1}}\)

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

= 1 - 1/ căn 100

=1 - 1/10

= 9/10

Ta có :

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

Ta có:

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

Do đó:

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)

\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)

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

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

\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

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

áp dụng vào tính ta được:

biểu thức cần tính: \(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}\)

\(=10-1=9\)

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

\(=-\sqrt{1}+\sqrt{100}\)

\(=-1+10=9\)

chúc bn học tốt