* ĐK: \(x\ne0\)

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

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

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

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

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

<=> \(\frac{x-1}{x\left(x+1\right)}=\frac{1}{36}\)

<=> \(\frac{x-1}{x\left(x-1\right)}=\frac{x-1}{36.\left(x-1\right)}\)

=> x(x-1) = 36. (x-1) => x =36

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

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

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

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

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

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

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

x+1=18

x=18-1

x=17

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

1/21 + 1/28 + 1/36 + ...+ 1/x(x+1)

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

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

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

=> 2. ( 1/6 - 1/7 + 1/7-1/8 + ...+ 1/x - 1/x+1

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

=>1/3 - 2/x+1

1/21 = 2/42 = 1/7-1/8

1/28 = 2/56 = 1/8-1/9

.....

1/(x*(x+1)) = 1/x -1/(x+1)

cộng lại với nhau ta sẽ được 1/7 - 1/(x+1) = 2/9

suy ra 1/(x+1) = -5/62 :D

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

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

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

=> \(\frac{A}{2}=\frac{1}{6}-\frac{1}{x+1}=\frac{x+1-6}{6\left(x+1\right)}=\frac{x-5}{6\left(x+1\right)}\) => \(A=\frac{x-5}{3\left(x+1\right)}=\frac{2}{9}\)

<=> 3(x-5)=2(x+1)  <=> 3x-15=2x+2  <=> x=17

Đáp số: x=17

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

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

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

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

\(\frac{1}{\left(x+1\right)}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\Leftrightarrow x+1=18\Leftrightarrow x=17\)

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

=>2/(6*7)+2/(7*8)+2/(8*9)+......+2/((x+(x+1))=2/9

=>1/6-1/7+1/7-1/8+1/8-1/9+....+1/x-1/x+1=2/9

=>1/6-1/x+1=2/9

=>1/x+1=1/6-2/9

Tự làm tiếp nhé 

\(17\)