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

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

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

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

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

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

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

\(\Leftrightarrow x+1=2011\)

\(\Leftrightarrow x=2010\)

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

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

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

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

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

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

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

\(x+1=2011\)

\(x=2011-1\)

\(\Rightarrow x=2010\)

 b,\(\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right):2=\frac{2013}{2015}:2\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)

\(\Rightarrow\)\(x+1=2015\)

\(\Rightarrow x=2014\)

a, 2/3x -3/2.x-1/2x=5/12

    x.(2/3-3/2-1/2)=5/12

                 x. -4/3=5/12

                          x=5/12:-4/3

                          x=-5/16

b,2/6+2/12+2/20+...+2/x.(x+1)=2013/2015

   2/2.3+2/3.4+2/4.5+...+2/x.(x+1)=2013/2015

   1/2(1-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015

                                                1/2(1-1/x+1)=2013/2015

                                                 1-1/x+1=2013/2015 : 1/2

                                                  1-1/x+1=4206/2015

                                                      suy ra đề sai

1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018

1/6+1/12+1/20+...+1/x(x+1)=504/1009

1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009

1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009

1/2-1/x+1=504/1009

x-1/2(x+1)=504/1009

-> 1009(x-1)=504.2(x+1)

1009x-1009=1008x+1008

1009x-1008x=1008+1009

->x=2017

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)

1/3 + 1/6 + 1/10 + .......+2/x(x + 1) =  2015/2017

=> 2/2.3 +2/3.4 + 2/4.5 +........+ 2/x(x+1) =2015/2017

=> 2. [1/2.3 + 1/3.4 +1/4.5+....+1/x(x+1) ] = 2015/2017

=> 2. [ 1/2+ (-1/3 + 1/3) + (-1 /4 +1/4)+ -1/5 +.......+ 1/x + -1/x+1]

=> (1/2 + -1/x+1) .2 =2015/2017

=> 1/2 + -1/x+1 = 2015/2017 :2 = 2015/2017 . 1/2 =2015/4034.

=>  -1/x+1 = 2015/4034 -1/2 = 2015/4034 -2017/4034 = -1/2017

=> -1/x+1 = -1/2017

=>x+1=2017

=> x= 2016

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

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

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

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

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

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

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

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

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

Mik mới lớp 5 thui à.  Sorry nha. 

\(\Rightarrow4x-\left(\frac{4}{5.7.9}+\frac{4}{7.9.11}+...+\frac{4}{99.101.103}\right)=\frac{2}{83224}=\frac{1}{41612}\)

\(4x-\left(\frac{9-5}{5.7.9}+\frac{11-7}{7.9.11}+...+\frac{103-99}{99.101.103}\right)=\frac{1}{41612}\)

\(4x-\left(\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)=\frac{1}{41612}\)

\(4x-\left(\frac{1}{5.7}-\frac{1}{101.103}\right)=\frac{1}{41612}\)

Từ đó tìm ra x

2/6+2/12+2/20+...+2/x.(x+1)=2013/2015

2.[1/6+1/12+1/20+...+1/x.(x+1)]=2013/2015

1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2013/4030

1/2-1/3+1/3-1/4+...+1/x-1/x+1=2013/4030

1/2-1/x+1=2013/4030

1/x+1=1/2015

=> x+1=2015

     x=2014

Vậy x=2014

Đặt A=Vế trái

Ta có :

\(A \over 2 \)\(= \)\({1\over 6 } +{1\over 12 }+{1\over 20 }+...+{1\over x(x+1)}\)

   =\({1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+{1\over4}-{1\over 5}+...+{1\over x-1}-{1\over x}+{1\over x}-{1\over x+1}\)

   =\({1\over2}-{1\over x+1}\)

Từ đó suy ra: \({1\over2}-{1\over x+1}={2013\over4030}\)

=> x=2014