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 :

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

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

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

Mỗi biểu thức trong dấu căn có dạng:

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\)   ( Với \(k\ge2\))

Ta có:

\(1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}\)

\(=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}\)

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

\(\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}\)

Mỗi biểu thức trong dấu căn có dạng:

1+1k2 +1(k+1)2    ( Với k≥2)

Ta có:

1+1k2 +1(k+1)2 =k2(k+1)2+(k+1)2+k2k2(k+1)2 =k4+2k3+k2+k2+2k+1+k2k2(k+1)2 

=k4+2k2(k+1)+(k+1)2k2(k+1)2 =(k2+k+1)2(k(k+1))2 

⇒√1+1k2 +1(k+1)2 =k2+k+1k2+k =1+1k(k+1) =1+1k −1k+1 

⇒S=1+1−12 +1+12 −13 +1+13 −14 +...+1+12013 −12014 =2014−12014 

Mỗi biểu thức trong dấu căn có dạng:

1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}1+k21​+(k+1)21​   ( Với k\ge2k≥2)

Ta có:

1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}=\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}=\frac{k^4+2k^3+k^2+k^2+2k+1+k^2}{k^2\left(k+1\right)^2}1+k21​+(k+1)21​=k2(k+1)2k2(k+1)2+(k+1)2+k2​=k2(k+1)2k4+2k3+k2+k2+2k+1+k2​

=\frac{k^4+2k^2\left(k+1\right)+\left(k+1\right)^2}{k^2\left(k+1\right)^2}=\frac{\left(k^2+k+1\right)^2}{\left(k\left(k+1\right)\right)^2}=k2(k+1)2k4+2k2(k+1)+(k+1)2​=(k(k+1))2(k2+k+1)2​

\Rightarrow\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\frac{k^2+k+1}{k^2+k}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}⇒1+k21​+(k+1)21​​=k2+kk2+k+1​=1+k(k+1)1​=1+k1​−k+11​

\Rightarrow S=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2013}-\frac{1}{2014}=2014-\frac{1}{2014}⇒S=1+1−21​+1+21​−31​+1+31​−41​+...+1+20131​−20141​=2014−

Với \(k\in N;k\ne0\) ta có :

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

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

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

Áp dụng ta có :

\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)

\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)

A = 

A = \(1-\frac{1}{2018}\)

A = \(\frac{2017}{2018}\)

Có : 

2.B = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

2.B = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

2.B = \(1-\frac{1}{2017}\)

2.B = \(\frac{2016}{2017}\)

B = \(\frac{2016}{2017}:2=\frac{1008}{2017}\)

Có :

3.C = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2017.2020}\)

3.C = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\)

3.C = \(\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

C = \(\frac{2019}{2020}:3=\frac{673}{2020}\)

a=1/1-1/2+1/2-1/3+...+1/2017-1/2018

=1/1-1/2018

=kq

may bai duoi lam tuong tu nha

mình chưa điền kết quả ban tu dien nha 

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2015\cdot2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(C=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2017\cdot2020}\)

\(C=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2017\cdot2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{2020}\right)\)

\(C=\frac{1}{3}\cdot\frac{2019}{2020}\)

\(C=\frac{673}{2020}\)

Xem lại đề

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

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

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

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