\(\frac{8}{1.5}+\frac{8}{5.9}+\frac{8}{9.13}+...+\frac{8}{x\left(x+4\right)}=\frac{1}{2}\)

\(\Leftrightarrow\)\(2\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{1}{2}\)

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

\(\Leftrightarrow\)\(1-\frac{1}{x+4}=\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{x+4-1}{x+4}=\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{x+3}{x+4}=\frac{1}{2}\)

\(\Rightarrow\)\(2\left(x+3\right)=x+4\)

\(\Leftrightarrow\)\(2x+6=x+4\)

\(\Leftrightarrow\)\(x=-2\)

Vậy....

P/s: tham khảo mk ko chắc là đúng

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

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

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

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

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

Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)

\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)

x + 4 = 113 ⇒ x = 109

\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)

Với \(A=\dfrac{56}{113}\) thì

 \(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)

\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=\dfrac{1}{113}\)

\(x=109\)

\(\dfrac{2}{1.5}\) + \(\dfrac{2}{5.9}\)  + \(\dfrac{2}{9.13}\)+...+  \(\dfrac{2}{x\left(x+4\right)}\)   = \(\dfrac{56}{113}\)

(\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+\dfrac{2}{9.13}+....+\dfrac{2}{x\left(x+4\right)}\) ) \(\times\) 2  = \(\dfrac{56}{113}\) \(\times\) 2

\(\dfrac{4}{1.5}\) + \(\dfrac{4}{5.9}\) +\(\dfrac{4}{9.13}\)+...+ \(\dfrac{4}{x\left(x+4\right)}\)                = \(\dfrac{112}{113}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

1  - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = 1 - \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = \(\dfrac{1}{113}\)

         \(x\) + 4 = 113

         \(x\)        = 113 - 4

         \(x\)        = 109

Đặt \(B=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+....+\frac{2}{97\cdot101}\)

\(\Rightarrow2B=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+....+\frac{4}{97\cdot101}\)

\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{101}\)

\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}\)

\(\Leftrightarrow B=\frac{96}{505}:2\)

Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)

\(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(A=\frac{1}{1}-\frac{1}{21}\)

\(A=\frac{20}{21}\)

\(\frac{20}{21}< 1\)

=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm ) 

* Mình sợ sai xD *

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

\(\text{Ta có :}\)

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

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

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

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

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

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

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

\(\text{Vậy ...}\)

           \(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

\(\dfrac{2}{1.5}\)+\(\dfrac{2}{5.9}\)+\(\dfrac{2}{9.13}\)+.................+\(\dfrac{2}{2013+2017}\)

=\(\dfrac{1}{1}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{13}\)+...................+\(\dfrac{1}{2013}-\dfrac{1}{2017}\)

=\(\dfrac{1}{1}-\dfrac{1}{2017}\)

=\(\dfrac{2017}{2017}+\dfrac{-1}{2017}\)

=\(\dfrac{2016}{2017}\)

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