Đặt: A= 1/3 +1/6+1/10+…+2/x(x+1)
A x 1/2 = 1/2.3 + 1/3.4 + 1/4.5 +…+1/x(x+1)
A x1/2 = 1/2-1/3+1/3-1/4+1/4-1/5+…..+1/x-1/(x+1)
A x 1/2 = 1/2 – 1/(x+1)
A = (1/2 -1/x+1) : 1/2
A = 1 – 2/(x+1)
Như vậy ta có: 1-2/(x+1) = 1999/2001
Hay: 2/(x+1) = 1-1999/2001
2/(x+1) = 2/2001
Vậy x = 2000

 Tích tớ nha!! Cáchgiải chính xác 100%

Ta có: 

1/3+1/6+1/10+...+1/x(x+1)=1999/2001

1/2.[1/3+1/6+1/10+...+1/x(x+1)].2=1999/2001

[1/6+1/12+1/20+...+1/x(x+1)].2=1999/2001

[1/2.3+1/3.4+1/4.5+...+1/x.(x+1)].2=1999/2001

[1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1].2=1999/2001

(1/2-1/x+1).2=1999/2001

1/2-1/x+1=1999/2001:2=1999/2001.1/2=1999/4002

1/x+1=1/2-1999/4002

1/x+1=2001/4002-1999/4002==2/4002=1/2001

=>x+1=2001

=>x=2001-1=2000

Vậy x=2000

Đặt A=1/3+1/6+1/10+...+2/x*(x+1)

        1/2A=1/3*2+1/6*2+1/10*2+...+2/2*x*(x+1)

         1/2A=1/6+1/12+1/20+...+1/x*(x+1)

          1/2A=1/2*3+1/3*4+1/4*5+...+1/x*(x+1)

           1/2A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)

           1/2A=1/2-1/x+1

           A=(1/2-1/x+1):1/2

          A=1-2/x+1

Ta có A=1999/2001

Hay 1-2/x+1=1999/2001

           2/x+1=1-1999/2001

          2/x+1=2/2001

=>x+1=2001

=>x=2000

Cho A = 1/3+1/6+1/10+...+2/x(x+1)

    1/2A= 1/3.2+1/6.2+1/10.2+...+2/x(x+1)2

    1/2A= 1/6+1/12+1/20+...+1/x(x+1)

    1/2A= 1/2.3+1/3.4+1/4.5+...+1/x(x+1)

    1/2A= 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1

    1/2A= 1/2-1/x+1

    A      = (1/2-1/x+1)/1/2

    A      = 1-2/x+1

Mà A=1999/2001

=> 1-2/x+1= 1999/2001

         2/x+1= 1-1999/2001

         2/x+1= 2/2001

     =>x+1=2001

     =>x     = 2000

Đặt N=1/10+1/15+1/21+...+2/x*(x+1)

1/2N=1/20+1/30+1/42+...+1/x*(x+1)

1/2N=1/4*5+1/5*6+1/6*7+...+1/x*(x+1)

1/2N=1/4-1/5+1/5-1/6+1/6-1/7+...+1/x-1/x+1

1/2N=1/4-1/x+1

N=(1/4-1/x+1):1/2

N=1/2-2/x+1

Thiếu đề

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}.\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)

=> x + 1 = 2001

=> x = 2010

\(\frac{1}{3}\)+  \(\frac{1}{6}\)+  \(\frac{1}{10}\)+  ..... +\(\frac{1}{X.\left(X+1\right)}\)=\(\frac{1999}{2001}\)

\(\frac{2}{2.3}\)+\(\frac{2}{2.6}\)+\(\frac{2}{2.10}\)+  ...... + \(\frac{1}{X.\left(X+1\right)}\)=\(\frac{1999}{2001}\)

\(\frac{2}{2.3}\)\(+\)\(\frac{2}{3.4}\)\(+\) \(\frac{2}{4.5}+...\) \(+\) \(\frac{1}{x\left(x+1\right)}\)=\(\frac{1999}{2001}\)

\(2\)\(.\)(\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\) ....) \(+\)\(\frac{1}{x\left(x+1\right)}\)\(=\)\(\frac{1999}{2001}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\)\(\frac{1999}{2001}:2\)

\(\frac{1}{2}-\frac{1}{x+1}\)\(=\frac{1999}{2001}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}\)

\(\frac{1}{x+1}=\frac{2}{4002}\)

\(\frac{1}{x+1}=\frac{1}{2001}\)

\(\Rightarrow x+1=2001\)

\(\Rightarrow x=2000\)

chúc bạn học giỏi. đúng thì k cho mình nha

quy luật gì vậy??

Tu de bai ta co 

1/6+1/12+1/20+...+1/(x*(X+1))=1999/4002

Suy ra 1/(2*3)+1/(3*4)+1/(4*5)+...+1/(x*(x+1))=1999/4002

Suy ra 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=1999/4002

Suy ra 1/2-1/(x+1)=1999/4002

Suy ra 1/(x+1)=1/2001

Suy ra x+1=2001

Suy ra x=2000

2000 do ban

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)

\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{199}{2001}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}\div2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\Leftrightarrow\frac{1}{x+1}=\frac{1}{2001}\)

\(\Leftrightarrow x+1=2001\Rightarrow x=2000\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{1999}{2001}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)

=> x + 1 = 2001

=> x = 2000

x=2000

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\)\(\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)

\(\Rightarrow x+1=2001\)

\(\Rightarrow x=2001-1=2000\)

Vậy \(x=2000.\)

Chỗ \(x\) phải là \(\frac{2}{x\left(x+1\right)}\) chứ bạn :) 

Ta có : 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\) ( nhân hai vế cho \(\frac{1}{2}\) ) 

\(\Leftrightarrow\)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(\Leftrightarrow\)\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{1999}{4002}\)

\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{4002}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2}-\frac{1999}{4002}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}=\frac{1}{2001}\)

\(\Leftrightarrow\)\(x-1=2001\)

\(\Leftrightarrow\)\(x=2001+1\)

\(\Leftrightarrow\)\(x=2002\)

Vậy \(x=2002\)

Chúc bạn học tốt ~ 

Sorry mình nhầm dưới mẫu nha bạn :') 

Sửa lại thành \(x+1\) nhé ~.~