\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2017}\)
\(\Leftrightarrow x+1=2017\)
\(\Leftrightarrow x=2017-1\)
\(\Leftrightarrow x=2016\)
Vậy x = 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\cdot\frac{x-1}{2\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{x-1}{2x+2}=\frac{2015}{4034}\)
\(\Rightarrow4034x-4034=4030x+4030\)
\(\Rightarrow4034x-4030x=8064\)
\(\Rightarrow x=2016\)