K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

6 tháng 4 2017

\(A=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2010^2}>1-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{2009.2010}\)

\(=1-\frac{1}{2}-\frac{1}{2010}=\frac{1004}{2010}>\frac{1}{2010}\Rightarrow A>\frac{1}{2010}\)

15 tháng 2 2017

\(\frac{B}{A}=\frac{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)

\(=\frac{\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+...+\left(\frac{1}{2012}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)

\(=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+....+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}}\)

\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}=2013\)

26 tháng 1 2017

\(A=1-\frac{1}{2^2}-...-\frac{1}{2010^2}\)

\(=1-\left(\frac{1}{2^2}+...+\frac{1}{2010^2}\right)\)

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\)

Ta có: \(A=1-\left(\frac{1}{2^2}+...+\frac{1}{2010^2}\right)\)\(>\)\(B=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)

\(=1-\left(1-\frac{1}{2010}\right)=1-1+\frac{1}{2010}=\frac{1}{2010}\)

26 tháng 1 2017

cảm ơn bn >.<!

bài bn vik thiếu nhưng mik hiểu nên vẫn tick

13 tháng 9 2016

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)

\(A=\frac{1}{2011}\)

 

13 tháng 9 2016

dunt

17 tháng 10 2018

\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)

\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)

\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)

\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)

Vậy \(\frac{B}{A}=2012\)

Chúc bạn học tốt ~ 

17 tháng 10 2018

cảm ơn bạn

17 tháng 2 2017

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)

\(=\frac{1}{2011}\)

6 tháng 2 2017

A= \(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)

B=\(\left(\frac{2012}{1}-1\right)+\left(\frac{2012}{2}-1\right)+...+\left(\frac{2012}{2011}-1\right)\)

= \(\frac{2012}{1}-\frac{2012}{2012}+\frac{2012}{2}-\frac{2012}{2012}+...+\frac{2012}{2011}-\frac{2012}{2012}\)

=\(2012\left(1-\frac{1}{2012}+\frac{1}{2}-\frac{1}{2012}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)

\(\Rightarrow\)\(\frac{B}{A}\)=\(\frac{2012\left(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)= 2012

23 tháng 10 2016

a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)

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

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

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

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

=>đpcm

b) Từ công thức trên ta có:

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

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

Ta có:

\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)

\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)

\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)