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

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

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(x=308-3\)

\(x=305\)

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

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

\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right) \)
                                                               \(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

 ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)

3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)

3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)

3A=\(\frac{1}{5}-\frac{1}{608}\)

3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)

Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)

      3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)

      3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)

      3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)

       A=\(\frac{201}{3040}\)

x=1

duyệt đi

\(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{65.68}\right)x=\frac{19}{68}+\frac{7}{34}\)

\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{68}\right)x=\frac{33}{68}\)

\(\left(\frac{1}{2}-\frac{1}{68}\right)x=\frac{33}{68}\)

\(\frac{33}{68}x=\frac{33}{68}\)

\(x=\frac{33}{68}:\frac{33}{68}=1\)

x=3

nhoc quay phá làm thiếu 

giúp với!!!

(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).x.3=33/68.3=99/68

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

(1/2-1/5+1/5-1/8+....-+1/65-1/68).x=99/68

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

33/68.x=99/68

=>x=3

k cho mình nha