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18 tháng 12 2016

lớn hơn , bé hơn hoặc bằng dễ òm đi chịch hk cưng ?

18 tháng 12 2016

ĐANG CẦN GẤP

18 tháng 2 2017

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

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 2 2017

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

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

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

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

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

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

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

14 tháng 1 2018

\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2011+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)

\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}}\)

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

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: \(P=\frac{1}{2012}\)