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

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

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

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

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

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

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

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

=>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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x=2004\)

Vậy x = 2004

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

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

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

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

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

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

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

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

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

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

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

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

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 ~ 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}=\frac{2005}{2007}\)

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

\(\Rightarrow1-\frac{2}{x+1}=\frac{2005}{2007}\)

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

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

\(\Rightarrow x+1=2007\)

\(\Rightarrow x=2006\)

\(\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{2005}{2007}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2005}{4014}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2005}{4014}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{4014}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{4014}\)

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

\(\Rightarrow x+1=2007\)

\(x=2007-1\)

\(x=2006\)