\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)

\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

Mik mới lớp 5 thui à.  Sorry nha. 

Viết sai đầu bài rồi!     

Ta có : 

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

\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)

\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)

\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)

Vậy \(x=\frac{2003}{2007}\)

Chúc bạn học tốt ~ 

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

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)

\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)

\(\Leftrightarrow4006x+4006=4010x-4010\)

\(\Leftrightarrow-4x=-8016\)

\(\Leftrightarrow x=2004\)

Vậy x = 2004

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

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

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)

\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)

\(\left(x+1\right).7=4020\)

\(\Rightarrow x=....\)

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

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

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

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2008\)

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

=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)

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

x+1=2009

x=2009-1=2008

Vậy x bằng 2008

x=2008

Ta có: 1/3+1/6+1/10+...+2/x*(x+1)

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

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

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

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

=2*(1/2-1/x+1)=2000/2002

=>1/2-1/x+1=2000/2002:2

=>1/2-1/x+1=500/1001

=>1/x+1=1/2-500/1001

=>1/x+1=1/2002

=>x+1=2002

=>x=2002-1

=>x=2001 thuộc N

Vậy x=2001

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