(1/2.5+1/5.8+....+1/65.68)x=19/68+7/34=33/68

(1/2.5+1/5.8+.....+1/65.68).3.x=33/68.3=99/68

(3/2.5+3/5.8+........3/65.68)x =99/68

(1/2-1/68)x=99/68

33/68x=99/68

=>x=3

cho mình nha

kfckfckfc ngon ngonngon

tich mik minh tich lai

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)

\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)

\(\Leftrightarrow10=-3x-9\)

\(\Leftrightarrow10+9=-3x\)

\(\Leftrightarrow19=-3x\)

\(\Leftrightarrow x=-\frac{19}{3}\)

Đề sai à -.- 

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)

=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)

=> \(10=-3\left(x+3\right)\)

=> 10 = -9x - 9

=> 10 + 9x + 9 = 0

=> 19 + 9x = 0

=> 9x = -19

=> x = -19/9

Nhầm khúc cuối 

=> 10 = -3x - 9

=> 10 + 3x + 9 = 0

=> 19 + 3x = 0

=> 3x = -19

=> x = -19/3

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

\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)

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

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

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

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)

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

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

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

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

=>x+3=308

<=>x=305 (nhận)

Vậy x=305

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

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

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

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

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\Rightarrow x+3=308\Rightarrow x=305\)

SAI ĐỀ!

Vớ vẩn đề đúng hẳn hoi , để tui làm cho coi nè.

Xét :\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)\)

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

Khi đó ,phương trình cần giải có dạng \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

Giờ thì việc tìm x quá đơn giản đúng không!

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

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

=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

=> \(x+3=308\)

=> x = 305

Vậy x = 305

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

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

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

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

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

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

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(\Leftrightarrow x=308-3\)

\(\Leftrightarrow x=305\)

Vậy \(x=305\)

Tìm x biết :

 a)  \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)(1)

Gọi B = \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\)

     \(\Rightarrow3A=3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)\)  

     \(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\)

     \(\Rightarrow3A=\frac{1}{5}-\frac{1}{x+3}\)

      \(\Rightarrow A=\left(\frac{1}{5}-\frac{1}{x+3}\right):3\)(2)

Thay (2) vào (1) , ta có : \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)

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

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

                                 \(\Rightarrow x+3=308\)

                                 \(\Rightarrow x=305\)

                                  Vậy x=305

Tương tự như câu b) nha 

Chúc bạn học tốt!! nhớ mik nha