K
Khách

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

12 tháng 11 2018

\(A=-\frac{1}{2010}-\left(\frac{1}{2010.2009}+\frac{1}{2009.2008}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

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

\(A=-\frac{1}{2010}-1+\frac{1}{2010}=-1\)

12 tháng 11 2018

\(A=-\frac{1}{2010}-\frac{1}{2010.2009}-\frac{1}{2009.2008}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=-\left(\frac{1}{2010}+\frac{1}{2010.2009}+\frac{1}{2009.2008}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

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

\(A=-1\)

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

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

Gọi \(S=\frac{2009}{1}+\frac{2008}{2}+...+\frac{1}{2009}\)

\(\Rightarrow S=\frac{2010-1}{1}+\frac{2010-2}{2}+...+\frac{2010-2009}{2009}\)

\(\Rightarrow S=2010-1+\frac{2010}{2}-1+...+\frac{2010}{2009}-1\)

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

\(\Rightarrow S=2010+\frac{2010}{2}+...+\frac{2010}{2009}-2009\)

\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+...+\frac{2010}{2009}+1\)

\(\Rightarrow S=\frac{2010}{2}+\frac{2010}{3}+..+\frac{2010}{2009}+\frac{2010}{2010}\)

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

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

18 tháng 2 2017

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

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

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

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

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

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

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

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

4 tháng 3 2018

nâng cao phát triển toán 7 đấy 

mấy bài đấu thì phải

4 tháng 3 2018

Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)

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

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

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

Hay: \(P=\frac{1}{2012}\)

5 tháng 11 2015

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

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

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

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

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