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

\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

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

\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=17\)

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

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

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

\(\Rightarrow2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

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

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

\(\Rightarrow x+1=18\Rightarrow x=17\)

Vậy x = 17

\(\Rightarrow\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+.....+\dfrac{2}{x\left(x+1\right)}\Rightarrow2\left(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+.....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}.\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{3}{54}\\ \Rightarrow x+1=\dfrac{54}{3}\\ \Rightarrow x=\dfrac{54}{3}-1=\dfrac{51}{3}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)

Làm tiếp :

​2. ( 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/ x + 1) = 2/9

​2. ( 1/6 - 1/ x + 1) = 2/9

​1/6 - 1/x + 1 = 2/9 : 2

​1/6 - 1/ x + 1 = 1/9

1/ x + 1 = 1/6 - 1/9

1 : x + 1 = 1/18

x + 1 = 18

x = 18- 1 = 17

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

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

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

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

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

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

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

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

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

\(\frac{1}{x-1}=\frac{1}{18}\)

\(\Rightarrow x-1=18\)

\(\Rightarrow x=18+1\)

\(\Rightarrow x=19\)

 = 2/42 + 2/56+2/72+................+2/x.(x+1)=2/9

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

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

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

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

=1/6-1/9=\(\frac{1}{x+1}\)=3/54=1/18

=> x= 18-1 =17

BẠN SKT_NTT sai đề mà

mình trả lời bài 1 thôi nhé :

Gọi biểu thức trên là A.

Theo bài ra ta có:A=1/1.6+1/6.11+1/11.16+...+1/(5n+1)+1/(5n+6)

                           =1/5(1-1/6+1/6-1/11+1/11-1/16+...+1/5n+1-1/5n+6)

                           =1/5(1-1/5n+6)

                           =1/5( 5n+6/5n+6-1/5n+6)

                           =1/5(5n+6-1/5n+6)

                           =1/5.5n+5/5n+6

                           =n+1/5n+6

                           =ĐIỀU PHẢI CHỨNG MINH

x- 20/11.13 - 20/13.15 - 20/13.15 - 20/15.17 -...- 20/53.55=3/11

x-10.(2/11.13+2/13.15+2/15.17+...+2/53.55=3/11

x-10.(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)=3/11

x-10.(1/11-1/55)=3/11

x-10.4/55=3/11

x-8/11=3/11

x = 3/11+8/11

x=11/11=1

****

ban Optimus Prime sai dau bai rui

Ta có : 

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\) ( cái đề hình như có 1 phân số \(\frac{2}{9}\) đúng không bạn ) 

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

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

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

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

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

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

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

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

\(\Leftrightarrow\)\(x+1=1:\frac{1}{18}\)

\(\Leftrightarrow\)\(x+1=18\)

\(\Leftrightarrow\)\(x=18-1\)

\(\Leftrightarrow\)\(x=17\)

Vậy \(x=17\)

Chúc bạn học tốt ~ 

fuck you

\(\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x.\left(x+1\right)}=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{30}+\frac{1}{42}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{x}\right)=\frac{806}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{806}{2015}:2\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{5}-\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=\frac{403}{2015}-\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

Chúc bạn học tốt !!!! 

\(\Rightarrow\frac{1}{2}\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{806}{2015}.\frac{1}{2}\)

\(\Rightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{403}{2015}\)

rồi bạn tự giải nốt nhé