\(=\frac{2014}{2015}\)

tk mk nha

Ta có : 

4/1 . 5 + 4/5 . 9 + ...+ 4/2001 . 2005 

= 1 - 1/5 + 1/5 - 1/9 + ...+ 1/2001 - 1/2005 

= 1 - 1/2005

=  2004/2005 

Tham khảo nha !!! 

lộn

\(=\frac{2004}{2005}\)

xl nha

\(S=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+..+\frac{2005-2001}{2001.2005}\)

\(=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+...+\left(\frac{1}{2001}-\frac{1}{2005}\right)\)

\(=1+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{9}+\frac{1}{9}\right)+...+\left(-\frac{1}{2001}+\frac{1}{2001}\right)-\frac{1}{2005}\)

\(=1-\frac{1}{2005}\)

\(=\frac{2004}{2005}\)

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{998.1000}\)

\(\Rightarrow B=3\left(\dfrac{1}{10.12}+\dfrac{1}{12.14}+...+\dfrac{1}{998.1000}\right)\)

\(\Rightarrow B=3.\dfrac{1}{2}\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{12}+...+\dfrac{1}{998}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}.\dfrac{99}{1000}\)

\(\Rightarrow B=\dfrac{297}{2000}\)

$\dfrac{4}{1.4}+\dfrac{5.9}+....+\dfrac{4}{2001.2005}$
$=1+\dfrac15-\dfrac19+....+\dfrac{1}{2001}-\dfrac{1}{2005}$
$=1-\dfrac{1}{2005}=\dfrac{2004}{2005}$

 \(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\)

\(=1+\dfrac{401}{2005}-\dfrac{1}{2005}\)

\(=1+\dfrac{400}{2005}=1+\dfrac{80}{401}=\dfrac{481}{401}\)

Giải:

\(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\) 

\(=1+\dfrac{80}{401}\) 

\(=\dfrac{481}{401}\)

B=\(\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{2001.2005}\)

=\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{2001}-\frac{1}{2005}\)

=\(\frac{1}{1}-\frac{1}{2005}\)

=\(\frac{2004}{2005}\)

\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)

\(=-\frac{1}{1}+\frac{1}{n}=\frac{1}{n}+1\)

bạn sửa số cuối tử là 4 nhé 

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}=1-\dfrac{1}{405}=\dfrac{404}{405}\)

\(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{401.405}\\ =1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{401}-\dfrac{1}{405}\\ =1-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{401}-\dfrac{1}{401}\right)-\dfrac{1}{405}\\ =1-0-0-....-0-\dfrac{1}{405}\\ =1-\dfrac{1}{405}\\ =\dfrac{404}{405}\)