= 2004 - 1 / 2003  + 2004

= 2003 / 2007

mình ko biết có đúng ko nữa         

2005 x 2004 - 1/2003 x 2005 + 2004

=2004 x ( 2005 + 1 ) - ...?

= 2005 x ( 2003 + 1 ) - 1/ 2003 x 2005 + 2004

= 2005 x 2003 + 2005 x 1 - 1/ 2003 x 2005 + 2004

= 2005 x 2003 + 2005 -1/ 2003 x 2005 + 2004

= 2003 x 2005 + 2004/ 2003 x 2005 + 2004

= 1 ( vì TS=MS)

Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)

\(=\dfrac{2006}{2007}\)

Tớ nghĩ là .................................................botay.kom

\(\frac{2005\times2004-1}{2003\times2005+2004}\)

=\(\frac{2005\times2004-1}{2003\times2005+\left(2005-1\right)}\)

=\(\frac{2005\times2004-1}{2003\times2005+2005-1}\)

=\(\frac{2005\times2004-1}{\left(2003+1\right)\times2005-1}\)

=\(\frac{2005\times2004-1}{2004\times2005-1}\)

=1

= 2005 x ( 2003 + 1 ) - 1/ 2003 x 2005 + 2004 

= 2005 x 2003 + 2005 x 1 - 1/ 2003 x 2005 + 2004

= 2005 x 2003 + 2005 -1/ 2003 x 2005 + 2004

= 2003 x 2005 + 2004/ 2003 x 2005 +2004

\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)

\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999+2004}\).

Có sai đề không vậy???

Sửa đề một chút :v

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999\cdot2004}\)

\(=\frac{1}{5}\left[\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{1999\cdot2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{1999}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\cdot\frac{125}{501}=\frac{25}{501}\)

Đặt A =1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004. Ta có:

        A= 1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004

      5A= 5/4 x 9 + 5/9 x 14 + 5/14 x 19 +...+ 5/1999 + 2004

      5A= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/1999 - 1/2004

      5A= 1/4 - 1/2004

        A= (1/4 - 1/2004)/5

13/276

52 tuần và 2 ngày

kết bạn với mình nha!

Năm 2004 có số tuần lễ và dư số ngày là

366:7=52 ( tuần lễ) dư 2 ngày

CAC BAN TRA LOI DUNG ROI

Có công thức \(\dfrac{x}{a\left(a+x\right)}=\dfrac{1}{a}-\dfrac{1}{a+x}\) nhé!

Ví dụ: \(\dfrac{2}{2.4}=\dfrac{1}{2}-\dfrac{1}{4}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)

Dấu . tức là nhân nhé!

\(\dfrac{2002\times2003+4004}{2005\times2004-4010}\\ =\dfrac{2002\times2003+2002\times2}{2005\times2004-2005\times2}\\ =\dfrac{2002\times2005}{2005\times2002}\\ =1\)

trả lời cho mình ik huhu☹