Ta có:

1/3 + 1/6 + 1/10 + ... + 1/x(x+1):2 = 2001/2003

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2001/2003

=> 2 [1/6 + 1/12 + 1/20 + ... + 1/x(x+1)] = 2001/2003

=> 2 [1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x+(x+1)] = 2001/2003

=> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1= 2001/2003 : 2

=> 1/2 - 1/x+1 = 2001/4006

=> 1/x+1 = 1/2 - 2001/4006 = 1/2003

=> x+1 = 2003 = 2002 + 1 

=>x = 2002

= 2/(2.3) + 2/3.4 + 2/4.5 +...+ 2/x(x+1) = 2 [1/2-1/3+1/3-1/4+...+1/x-1/(x+1)]

=2[1/2-1/(x+1)]= (x-1)/(x+1) = 2001/2003

==> x=2002

x=2002

Câu hỏi 

1/3+1/6+1/10+...+1/x(x+1):2=2001/2003

=Tôi ko biết nên tôi xin hết 

Tron vn

mik nghĩ chỗ \(\dfrac{2}{x.\left(x+1\right)}\) phải là \(\dfrac{1}{x.\left(x+1\right)}\) bạn có thể vui lòng kiểm tra lại đề không Lệ Quyên

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

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)

\(\Leftrightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)

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

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2003}\)

\(\Leftrightarrow x+1=2003\Leftrightarrow x=2002\)

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

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)

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

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

=> x+1 = 2003

=> x = 2003 - 1

=> x = 2002

thank you very much