\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{n.\left(n+3\right)}\)=\(\frac{1}{3}\)(\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{n}+\frac{1}{n+3}\))=\(\frac{1}{5}-\frac{1}{n+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)

=>\(\frac{1}{n+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)vậy n= 308+3=311

Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^

duyệt đi

Đơn giản -_-

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)

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

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

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

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

\(\Rightarrow\)x=305

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

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

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

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

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

=>x+3=308

x=308-3

x=305

Vậy x=305

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

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

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

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

\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)

b ko hiểu đề

       Câu hỏi của Nguyễn Hữu Hưng       

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(x=305\)

\(a;343-4\times\left(x+1\right)=143\)

\(4\times\left(x+1\right)=343-143\)

\(4\times\left(x+1\right)=200\)

\(x+1=200:4\)

\(x+1=50\)

\(x=50-1=49\)

Mấy câu hỏi trẻ trâu thì tự tính nhé

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

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

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

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

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

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

\(\Leftrightarrow x+3=308\)

\(\Leftrightarrow x=305\)

Vậy x=305

\(131+\left(26-3\times x\right):5=135\)

\(\left(26-3\times x\right):5=135-131\)

\(\left(26-3\times x\right):5=4\)

\(26-3\times x=4\times5\)

\(26-3\times x=20\)

\(3\times x=26-20\)

\(3\times x=6\)

\(x=6:3=2\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\Rightarrow\frac{x-2}{5x+15}=\frac{303}{1540}\)

\(\Rightarrow x=305\)

x = 305 nha bạn . 

Gio thi ban may man trong ca thang 

Nguyen Dinh Tung

Nguen Quynh Trang se vui ve trong toi nay