\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(=\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\frac{1}{n+1}=\frac{1}{4008}\)

\(\Rightarrow\)n+1=4008

n=4007

Vậy n=4007

 TA CÓ :\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{n\left(n+1\right)}\)\(=\frac{2003}{2004}\)

      \(Nhân\)\(cả\)\(hai\)\(vế\)\(với\)\(\frac{1}{2}\),  TA ĐƯỢC :

    \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n.\left(n+1\right)}\right)\)\(=\frac{1}{2}.\frac{2003}{2004}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{n.\left(n+1\right)}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n.\left(n+1\right)}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{n}-\frac{1}{n+1}\)\(=\frac{2003}{4008}\)

=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

=>\(\frac{1}{n+1}=\frac{1}{4008}\)

=> \(n+1=4008\)

=> \(n=4007\)( Thỏa mãn điều kiện : \(n\in N\))

Vậy n=4007 

1/3 + 1/6 + 1/10 + ... + 2/n(n+1) = 2003/2004

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)

\(\Rightarrow n+1=4008\)

\(\Rightarrow n=4008-1=4007\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{2003}{2004}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}=\frac{1}{4008}\)

\(\Rightarrow n+1=4008\Rightarrow n=4007\)

Vậy \(n=4007\)

đặt a=1/3+1/6+1/10+...........+2/n(n+1)

1/2a=1/6+1/12+...........+1/n(n+1)

1/2a=1/2.3+1/3.4+........+1/n(n+1)

1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1

1/2a=1/2-1/n+1

a=(1/2--1/n+1):1/2=2003/2004

1/2-1/n+1=2003/2004.1/2

1/2-1/n+1=2003/4008

1/n+1=1/2-2003/4008

1/n+1=1/4008

suy ra n+1=4008

n=4007

n=4007 do

n=4007

Để \(\frac{4x+9}{6x+5}\)\(\in Z\)thì \(4x+9\)chia hết \(6x+5\)

                                \(\Rightarrow3.\left(4x+9\right)\)chia hết cho \(6x+5\)

                                \(\Rightarrow\)\(12x+27\)chia hết cho \(6x+5\)

                                \(\Rightarrow\)\(2.\left(6x+5\right)+17\)chia hết cho \(6x+5\)

                                \(\Rightarrow\)17 chia hết cho \(6x+5\)

                                 \(\Rightarrow\)6x +5 thuộc Ư(17)

                                  suy ra 6x+5 thuộc {+-1;+-17}

                       ĐẾN ĐÂY BẠN TỰ LẬP BẲNG TÌM X NHÉ

       Vậy x thuộc{-1;2}

B)Tích đi mình làm tiếp cho

Có: 1/3+1/6+1/10+...+2/n(n+1)=2003/2004

 =>1/2.[ 1/3+1/6+1/10+...+2/n(n+1)]=2003/2004.1/2

=>1/6+1/12+1/20+...+1/n.(n+1)=2003/2004.1/2

=>1/2.3+1/3.4+1/4.5+...+1/n.(n+1)=2003/2004.1/2

=>1/2-1/3+1/3-1/4+1/4-1/5+....+1/n-1/n+1=2003/2004.1/2

=>1/2-1/n+1=2003/4008

=>1/n+1=1/4008

=>n+1=4008

=>n=4007

Vậy n=4007

Tích đi bạn mình làm cho 2 câu rồi đấy