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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=17\)

Đề sai nha

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

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

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

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

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

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

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

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

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

\(\Rightarrow x+1=18\)

Tự làm tiếp nha!

Đúng ời. đề mk viết lộn chút xíu

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

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

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

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

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

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

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

<=> \(\frac{1}{x+1}=\frac{1}{18}\)

<=> x + 1 = 18 

=> x = 18 - 1 

x  = 17 

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

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

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

Làm ơn mai mốt có viết bài gì viết phân số cho dễ coi nhé !

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

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)= \(\frac{2}{9}\)

\(\frac{2}{9}\) - \(\frac{1}{9}\)= \(\frac{1}{9}\)

x.(x + 1 ) = 9 ( bạn nên kiểm lại đi vì có lẽ bạn sai đề

Theo mình lập luận mãi thì 2.(2 + 1 ) = 6 mà 3 ( 3 + 1 ) = 12 cho nên bạn nên kiểm lại đề, đồng thời cũng nên kiểm lại đề của mình bằng cách trình bày theo phân số rồi mik sẽ giải tiếp.