Giải:

Ta có công thức sau:

\(\frac{k}{a.b}=\frac{1}{a}-\frac{1}{b}\) với b - a = k hoặc a - b = k

Lắp vào biểu thức A, ta có:

\(A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.14}+...+\frac{4}{2005.2009}\\ =\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{2001}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2009}\)

\(=1+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2009}\\ =1-\frac{1}{2009}\\ =\frac{2009-1}{2009}\\ =\frac{2008}{2009}\)

Vậy \(A=\frac{2008}{2009}\)

Chúc bạn học tốt!

Thanks bạn nha!

\(\text{Đề bài sai : }\frac{4}{\left(n-4\right)^n}->\frac{4}{\left(n-4\right)^n}\)

\(\text{Ta có :}\)

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

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

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

                                                  \(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)

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

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

                                                  \(=\frac{3n+5}{5n}\)

\(\text{Vậy ...}\)

Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)

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

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

\(=-\left(1-\frac{1}{n+4}\right)\)

\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)

\(=-\frac{n+3}{n+4}\)

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

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

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

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

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

\(M=1-\frac{1}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(M=1-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)

\(M=\frac{3}{5}+\frac{1}{n}\)

Mình chỉ giải đến đây thôi vì chẳng biết n bằng mấy cả

= - (1-1/5 +1/5 -1/9 +1/9 -1/13 +1/n + 1/n+4)

=-(1-1/n+4)

=-1+1/n+4

M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n

M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n

M = -(1-1/n)

M = -1 + 1/n 

M = -n + 1

xie xie bn nha

a) \(x+\)\(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)

\(\Rightarrow x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}=-\frac{4}{5}\)

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

b) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)

\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2A=1-\frac{1}{2005}\)

\(\Rightarrow2A=\frac{2004}{2005}\)

\(\Rightarrow A=\frac{1002}{2005}\)

Tính tổng:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\) 

= \(\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\right)\)  

= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2003}-\frac{1}{2005}\right)\) 

= \(\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

= \(\frac{1}{2}\cdot\frac{2004}{2005}\)  

= \(\frac{1002}{2005}\) 

k nha