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

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

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

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

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

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

=> x+1 =18

=> x = 18 - 1

=> x = 17

giúp với

sorry nha ko biết làm

ta có:1/21+1/28+1/36+...+1/x.(x+1)=2/9

=>1/(3.7)+1/(4.7)+1/(4.9)+...+2/x(x+1)=2/9

=>2/(6.7)+2/(7.8)+2/(8.9)+...+2/x(x+1)=2/9

=>2.[1/6-1/(x+1)]=2/9

=>1/6-1/(x+1)=1/9

=>1/(x+1)=1/18

=>x+1=18

=>x=17.

Vậy x = 17.

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

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

\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

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

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

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}=\frac{1}{18}\)

\(\Rightarrow x+1=18\)

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

Vậy \(x=17\)

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x-1=18\)

\(\Rightarrow x=18+1\)

\(\Rightarrow x=19\)

 = 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

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