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

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)

\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)

=>x+1=2005

=>x=2004

1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015

Đ/A là 2004

chúc đồng chí Chế Minh Hải học tốt

2.

a) Ta có:

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)

Vậy x = -1

b) Ta có:

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)

Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)

Vậy, x = -2004

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

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha

\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)

\(\Leftrightarrow x+1=2011\)

\(\Leftrightarrow x=2010\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(1-\frac{2}{x+1}=\frac{2009}{2011}\)

\(\frac{2}{x+1}=1-\frac{2009}{2011}\)

\(\frac{2}{x+1}=\frac{2}{2011}\)

\(x+1=2011\)

\(x=2011-1\)

\(\Rightarrow x=2010\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)

\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)

\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)

\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)

\(1-\frac{2}{x+1}=\frac{2007}{2009}\)

\(\frac{2}{x+1}=1-\frac{2007}{2009}\)

\(\frac{2}{x+1}=\frac{2}{2009}\)

\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)