Ta có :

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

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

\(=\frac{2017036}{2009}=1004\)

Ta có ; \(\frac{1}{2009}+\frac{2}{2009}+\frac{3}{2009}+......+\frac{2008}{2009}\)

\(=\frac{1+2+3+......+2008}{2009}\)

\(=\frac{2017036}{2009}=1004\)

Ta có :

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

 \(=\frac{1+2+3+....+2008}{2009}\)

\(=\frac{2017036}{2009}=1004\)

A/ \(\left(10\frac{3}{4}+3\frac{4}{5}\right)-\left(5\frac{3}{4}-1\frac{1}{5}\right)\)

\(=\left(10\frac{3}{4}-5\frac{3}{4}\right)+\left(3\frac{4}{5}+1\frac{1}{5}\right)\)

\(=5+5\)

\(=10\)

chúc bạn học tốt nha

A = 3 + 6 + 9 + ... + 2007 

=>A = 3( 1 + 2 + 3 + ... + 669 )

=> A = \(3\cdot\left(\frac{670\cdot669}{2}\right)\)

=> A = \(3\cdot224115\)= 672345

B = \(2\cdot53\cdot12+4\cdot6\cdot87-3\cdot8\cdot40\)

=> B = 24 * 53 + 24 * 87 - 24 * 40

=> B = 24 * ( 53 + 87 - 40 )

=> B = 24 * 100 = 2400

c) ta có Tử số = \(2006\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)\)

Mẫu số = \(\frac{2007-1}{1}\)+\(\frac{2007-2}{2}\)+...+\(\frac{2007-2006}{2006}\)

=> Mẫu số = \(\frac{2007}{1}\)\(-1\)+ \(\frac{2007}{2}\)\(-1\)+ ... + \(\frac{2007}{2006}\)\(-1\)

=> Mẫu số = \(\frac{2007}{1}\)+ \(\frac{2007}{2}\)+ ... + \(\frac{2007}{2006}\)- ( 1 + 1 + 1 + ... + 1 )        ( 1 + 1 + ... + 1  có 2006 số hạng 1 )

=> Mẫu số =  ( 2007 - 2006 ) + \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)

=> Mẫu số = \(\frac{2007}{2007}\)+ \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)

=> Mẫu số = \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)\)

=> C = \(\frac{TS}{MS}\)= \(\frac{2006}{2007}\)

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

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

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

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

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

\(\Rightarrow D=\frac{1}{2012}\)

Xét tử ta có:

\(2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{1}{2008}\)

= \(1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)\)

= \(\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)

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

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

=> A = 2009

A=\(\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...........+\left(1+\frac{2}{2008}\right)+\left(1+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2008}+\frac{1}{2009}}\)=\(\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+....+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)  

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

                                                                                                                =2009 

Vay A=2009

A= 2009

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

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

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

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

\(\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}\)

\(=\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}.\frac{1}{7}\)

\(=\frac{-3}{5}\left(\frac{4}{7}+\frac{2}{7}+\frac{1}{7}\right)\)

\(=\frac{-3}{5}.1\)

\(=\frac{-3}{5}\)

\(-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{2}{7}+-\frac{3}{5}=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}\left(\frac{2}{7}+1\right)=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{9}{7}.\)

\(-\frac{2}{35}-\frac{27}{35}=-\frac{29}{35}\)

tôi tính nhẩm vậy ko biết có đứng ko

Bạn CTV viết sai đề rồi kìa.

\(C=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}}\)

Đặt N = \(\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}\)

\(\Rightarrow N=\frac{1}{2006}+.....+\frac{2004}{3}+\frac{2005}{2}+\frac{2006}{1}\)

\(\Rightarrow N=\left(\frac{1}{2006}+1\right)+.....+\left(\frac{2004}{3}+1\right)+\left(\frac{2005}{2}+1\right)+1\)( Có 2005 nhóm )

\(=\frac{2007}{2006}+....+\frac{2007}{3}+\frac{2007}{2}+\frac{2007}{2007}\)
\(=2007\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2006}+\frac{1}{2007}\right)\)

Đặt M = \(\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}\)

\(=2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)\)

Thay N và M vào C , ta có :

\(C=\frac{N}{M}=\frac{2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)}{2007\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}\right)}=\frac{2006}{2007}\)

\(\Rightarrow C=\frac{2006}{2007}\)

Vậy : \(C=\frac{2006}{2007}\)