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

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

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

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

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

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

=> 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

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)

\(\Rightarrow x+1=18\)

\(\Rightarrow x=17\)

câu a khó quá.Để nghĩ.

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)

\(\Rightarrow9\left(x-5\right)=6x+6\)

\(\Rightarrow9x-45=6x+6\)

\(\Rightarrow9x-6x=51\)

\(\Rightarrow3x=51\)

Tới đây bí:v

\(\frac{2}{3}x-\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Rightarrow\frac{2}{3}x-\frac{3}{2}x+\frac{3}{4}=\frac{5}{12}\)

\(\Rightarrow\left(\frac{2}{3}x-\frac{3}{2}x\right)+\frac{3}{4}=\frac{5}{12}\)

\(\Rightarrow\frac{-5}{6}x+\frac{3}{4}=\frac{5}{12}\)

\(\Rightarrow\frac{-5}{6}x=\frac{5}{12}-\frac{3}{4}\)

\(\Rightarrow\frac{-5}{6}x=\frac{5}{12}-\frac{9}{12}\)

\(\Rightarrow\frac{-5}{6}x=\frac{-4}{12}\)

\(\Rightarrow\frac{-5}{6}x=\frac{-1}{3}\)

\(\Rightarrow x=\frac{-1}{3}:\frac{-5}{6}\)

\(\Rightarrow x=\frac{2}{5}\)

a)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+2}=\frac{2}{9}:2\)

\(\frac{1}{x+2}=\frac{1}{6}-\frac{1}{9}\)

\(\frac{1}{x+2}=\frac{1}{18}\)

=>x+2=18

=>x=16

b tương tự nhân nó với 1/2

Cám ơn bạn

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 

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