Giải:

Ta có:

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)

De thoi ma,minh ko ghi lai de nha

=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+...+...=\(\frac{1}{6}\)

=\(\frac{1}{2}\)-\(\frac{1}{x}\)=\(\frac{1}{6}\)

Con lai bn tu lam nha . chuc bn hok tot !!!

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+....+\frac{3}{\left(x-3\right).x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}\)

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

\(\Leftrightarrow x=1\div\frac{1}{3}=3\)

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (dấu . là nhân nhé)

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

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

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)

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

=> \(x+3=308\Rightarrow x=308-3=305\)

\(\frac{1}{5x8}+\frac{1}{8x11}+...+\frac{1}{Xx\left(X+3\right)}=\frac{101}{1540}\)

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

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

<=>\(\frac{x-2}{5x+15}=\frac{101}{1540}\)

<=>1540x-3080=505x+1515

<=>1035x=4595

<=>x=919/207

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

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{7}{14}-\frac{1}{14}\)

\(=\frac{6}{14}\)

\(=\frac{3}{7}\)

3/2x5 + 3/5x8 + 3/8x11 + 3/11x14 

= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14 

= 3/2 - 3/14 

= 21/14 - 3/14 

= 18/14 

= 9/5 

\(B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

toán 6 hả

X3 lên nhé

X3 lên

nhân cả tử và mẫu cho 3 là làm đc thôi

lên mạng mà tra

Đặt \(Shin=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)

\(\Rightarrow3Shin=\frac{6}{2}-\frac{6}{5}+\frac{6}{5}-\frac{6}{8}+\frac{6}{8}-\frac{6}{11}+...+\frac{6}{98}-\frac{6}{101}\)

\(\Leftrightarrow3Shin=\frac{6}{2}-\frac{6}{101}=\frac{297}{101}\)

N/t: . là dấu nhân nha! Cái đó lớp 5 chưa biết đâu! Lên cấp 2 mới học. Trong bài làm bạn cứ ghi là dấu  " x"  thay cho dấu "." của mình nha!

Đặt A = 6/2.5 + 6/5.8 + ... + 6/98.101

=> A = 6.(1/2.5 + 1/5.8 + ... + 1/98.101)

=> 3A = 6.(3/2.5 + 3/5.8 + ... + 3/98.101)

=> 3A = 6.(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/98 - 1/101)

=> 3A = 6.99/202

=> 3A = 297/101

=> A = 99/101

x là nhân hả

\(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)

\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{98.101}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{98}-\frac{1}{101}\right)\)

\(=2\left(\frac{1}{2}+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{8}+\frac{1}{8}\right)+....+\left(\frac{-1}{98}+\frac{1}{98}\right)-\frac{1}{101}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=2.\frac{99}{202}\)

\(=\frac{99}{101}\)

\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)

\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

_Chúc bạn học tốt_

Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)

\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)

\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)

\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)

\(3xA=\frac{1}{2}-\frac{1}{20}\)

\(3xA=\frac{9}{20}\)

\(\Rightarrow A=\frac{3}{20}\)