Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:

a) Ta có

\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)

\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)

\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)

\(=\sqrt{2017}-1\)

\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)

b) Ta có

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

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

\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Tương tự ta có

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

......................

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

Suy ra

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

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

a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5  + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \frac{{(\sqrt 5  + 1)(\sqrt 5  - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017}  + \sqrt {2013} )(\sqrt {2017}  - \sqrt {2013} )}}{{\sqrt {2013}  + \sqrt {2017} }}\\
 = \sqrt 5  - 1 + \sqrt 9  - \sqrt 5  + ... + \sqrt {2017}  - \sqrt {2013} \\
 = 1 + \sqrt 5  - \sqrt 5  + \sqrt 9  - \sqrt 9  + ... + \sqrt {2013}  - \sqrt {2013}  + \sqrt {2017} \\
 = 1 + \sqrt {2017} \\
 \Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]

Cảm ơn Tran Tu, kết bn đê

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

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

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

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

Áp dụng: \(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)

Ta có:

\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2.\left(n+1\right)^2}}\)

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

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

Thế vào bài toán ta được

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

\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)  

Bạn hãy chứng minh đẳng thức phụ sau : \(\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|1-\frac{1}{k}+\frac{1}{k+1}\right|\)

Áp dụng : \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\left(1+1-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)\(=1.99+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100-\frac{1}{100}\)

Với a \(\in\)N*, ta có:

\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\frac{a^2.\left(a+1\right)^2}{a^2.\left(a+1\right)^2}+\frac{\left(a+1\right)^2}{a^2.\left(a+1\right)^2}+\frac{a^2}{a^2.\left(a+1\right)^2}}\)

\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+\left(a+1\right)^2+a^2}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+a^2+2a+1+a^2}{\left[a.\left(a+1\right)\right]^2}}\)

\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2a^2+2a+1}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2.\left(a^2+a\right)+1}{\left[a.\left(a+1\right)\right]^2}}\)

\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2.a.\left(a+1\right).1+1^2}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)+1\right]^2}{\left[a.\left(a+1\right)\right]^2}}\)

\(=\sqrt{\left[\frac{a.\left(a+1\right)+1}{a.\left(a+1\right)}\right]^2}=\frac{a.\left(a+1\right)+1}{a.\left(a+1\right)}=\frac{a.\left(a+1\right)}{a.\left(a+1\right)}+\frac{1}{a.\left(a+1\right)}\)

\(=1+\frac{a+1-a}{a.\left(a+1\right)}=1+\frac{a+1}{a.\left(a+1\right)}-\frac{a}{a.\left(a+1\right)}=a+\frac{1}{a}-\frac{1}{a+1}\)

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

Thay a=1,2,...99

=>\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+1-\frac{1}{2}\)

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3}\)

............................................................

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

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

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

\(=1+1+...+1-\frac{1}{100}\)

\(=100-\frac{1}{100}\)

\(=\frac{9999}{100}\)

Vậy \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\frac{9999}{100}\)

Mình xin lỗi nhé, đẳng thức phụ phải là : \(\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|1+\frac{1}{k-1}-\frac{1}{k}\right|\)

Với mọi n thuộc N ta có :

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

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

Áp dụng ta được :

\(S=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+....+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)

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

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

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

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