Ta có : 

\(A=\frac{1}{8.11}+\frac{1}{11.14}+..+\frac{1}{605.608}\)

\(3A=\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)

\(3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)

\(3A=\frac{1}{8}-\frac{1}{608}\)

\(A=\frac{75}{608}:3\)

\(A=\frac{25}{608}\)

Ủng hộ mk nha !!! ^_^

Ta có: \(A=\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{605.608}\)

\(\Rightarrow3A=\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{605.608}\)

\(\Rightarrow3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{605}-\frac{1}{608}\)

\(\Rightarrow3A=\frac{1}{8}-\frac{1}{608}=\frac{75}{608}\)

\(\Rightarrow A=\frac{75}{608}:3=\frac{25}{608}\)

Vậy \(A=\frac{25}{608}\)

Ủng hộ cho mik nha bạn?

Ta có : 

A=18.11 +111.14 +..+1605.608 

3A=38.11 +311.14 +...+3605.608 

3A=18 −111 +111 −114 +...+1605 −1608 

3A=18 −1608 

A=75608 :3

A=25608 

25/608 tick nha cho tròn 60

A = \(\frac{1}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+.......+\frac{3}{605.608}\right)\)

   = \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{605}-\frac{1}{608}\right)\)

   = \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{608}\right)=\frac{1}{3}.\frac{76-1}{608}=\frac{1}{3}.\frac{75}{608}=\frac{25}{608}\)

\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{97.100}\)

\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{97.100}\right)\)

\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}.\frac{49}{100}=\frac{49}{300}\)

Ta có: \(S=\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{97.100}.\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{97.100}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Rightarrow S=\frac{49}{100}:3=\frac{49}{300}\)

Vậy \(S=\frac{49}{300}\)

CHÚC BẠN HỌC TỐT

\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\)

\(S=3\cdot\frac{1}{3}\cdot\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{97\cdot100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{50}{100}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\frac{49}{100}\)

\(S=\frac{49}{300}\)

\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(F=\frac{1}{5}-\frac{1}{2009}\)

\(F=\frac{2004}{10045}\)

\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{1}{11.14}+...+\frac{3}{2006.2009}\)

\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(F=\frac{1}{5}-\frac{1}{2009}\)

\(F=0\)

F = 1/5 - 1/8 + 1/8 - 1/11 +,,,+1/2006 - 1/2009
F = 1/5 - 1/2009
F = 2008/10044
Chúc bạn học tốt 

a)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540

<=>1/3(3/5.8+3/8.11+...+3/x(x+3)     =101/1540

<=>1/3(1/5-1/8+1/8-1/11+...+1/x-1/x+3=101/1540

<=>1/5-1/x+3=303/1540<=>1/x+3=1/308

<=>x+3=308<=>x=305

Nguồn CHTT, hihi !

Tham gia event này đi mọi người https://olm.vn/hoi-dap/detail/227766827875.html

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

Tính

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{17}{34}-\frac{2}{34}=\frac{15}{34}\)

Ta có: Biểu thức trên sẽ =

\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)1/17 (nó bị lỗi nên k viết đc T_T)

= \(\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

Chúc bạn học tốt!~

đề sai ak bn

mình sửa lại đề nhé

        \(\frac{1}{5.8}\) +    \(\frac{1}{8.11}\)  +     \(\frac{1}{11.14}\)    + ... +    \(\frac{1}{x.\left(x+3\right)}\) = \(\frac{101}{1540}\) 

\(\Rightarrow\)\(\frac{1}{3}\)(\(\frac{3}{5.8}\)+    \(\frac{3}{8.11}\) +   \(\frac{3}{11.14}\)  +  ...  +    \(\frac{3}{x.\left(x+3\right)}\) =  \(\frac{101}{1540}\)

\(\Rightarrow\) \(\frac{1}{3}\)( \(\frac{1}{5}\) - \(\frac{1}{8}\) +\(\frac{1}{8}\)-\(\frac{1}{11}\)+...+\(\frac{1}{x}\)- \(\frac{1}{x+3}\)) = \(\frac{101}{1540}\)

\(\Rightarrow\)  \(\frac{1}{3}\)(\(\frac{1}{5}\)- \(\frac{1}{x+3}\)) = \(\frac{101}{1540}\)

\(\Rightarrow\)    \(\frac{1}{5}\)-  \(\frac{1}{x+3}\)  =  \(\frac{101}{1540}\):   \(\frac{1}{3}\)

\(\Rightarrow\)    \(\frac{1}{5}\) -  \(\frac{1}{x+3}\)= \(\frac{303}{1540}\)

\(\Rightarrow\) \(\frac{1}{x+3}\)   =  \(\frac{1}{5}\)-  \(\frac{303}{1540}\)

\(\Rightarrow\) \(\frac{1}{x+3}\)=  \(\frac{1}{308}\)

\(\Rightarrow\) x+3 = 308

\(\Rightarrow\)x        = 305

Vậy x = 305

      \(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)

\(\Leftrightarrow x+3=103\)

\(\Leftrightarrow x\)\(=103-3\)

\(\Leftrightarrow x\)\(=100\)

Vậy x = 100

~~~~~~~Hok tốt~~~~~~~~

ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)

\(\Rightarrow x+3=103\)

\(\Rightarrow x=100\)

nhớ k nha

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14+}+...+\frac{1}{x\left(x+3+\right)}=\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{3\left(x+3\right)}=3.\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)

\(\Leftrightarrow x+3=103\)

\(\Leftrightarrow x+103-3\)

\(\Leftrightarrow x=100\)

vậy x=100

~~~~~~~~~~~~~~~~~~~~hk tốt~~~~~~~~~~~~~~~~~~~~

a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)

=> x + 3 = 308

     x = 308 - 3

     x = 305

b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)

=> x + 1 = 1993

     x = 1993 - 1

     x = 1992

a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(x=308-3\)

\(x=305\)