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

Phạm Thị Mai Anh ko có dấu( - )đằng trước đâu ạ

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

\(\text{Ta có :}\)

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

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

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

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

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

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

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

\(\text{Vậy ...}\)

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

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

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

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

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

Ta có:

\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)

b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)

\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)

\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)

\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)

Bn Tấn sai rùi

phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)

Ta có: \(\frac{1}{20.21}=\frac{21-20}{20.21}=\frac{21}{20.21}-\frac{20}{20.21}=\frac{1}{20}-\frac{1}{21}\)

Suy ra: \(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{20}-\frac{1}{61}=\frac{41}{1220}\)

Phép tính sau làm tương tự.

\(D=4\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{201.205}\right)\)
\(D=4\left(\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\left(\frac{1}{201}-\frac{1}{205}\right)\right)\)
D=4[(1-1/205)
D=4.204/205
=>D=816/205
____________________--
li-ke cho mình nhé bn Cao Minh Hoàng

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

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

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

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

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

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

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

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{9}{5.9}-\frac{5}{5.9}+\frac{13}{9.13}-\frac{9}{9.13}+\frac{17}{13.17}-\frac{13}{13.17}+...+\frac{45}{41.45}-\frac{41}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{21}{45}\)

\(\Rightarrow\frac{21}{3x}=\frac{21}{45}\)

\(\Rightarrow3x=45\)

\(\Rightarrow x=15\)