\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{y.\left(y+1\right)}=\frac{2008}{4002}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{y}-\frac{1}{y+1}\right)=\frac{1999}{2002}\)
\(=2.\left(\frac{1}{2}-\frac{1}{y+1}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{y+1}=\frac{1999}{2001}:2\)
\(\frac{1}{y+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{y+1}=\frac{1}{2001}\)
\(\Rightarrow y+1=2001\Rightarrow y=2000\)
Ta có: 1/3+1/6+1/10+...+2/y×(y+1)=1999/2001
2/6+2/12+2/20+...+2/y×(y+1)=1999/2001
2×(1/2×3+1/3×4+1/4×5+...+2/y×(y+1)=1999/2001
2×(1/2-1/3+1/3-1/4+1/4-1/5+...+1/y-1/y+1=1999/2001
(1/2-1/y+1)=1999/2001 : 2
1/2-1/y+1=1999/4002
1/y+1=1/2-1999/4002
1/y+1=1/2001
=>y+1=2001
=>y=2001
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{y.\left(y+1\right)}=\frac{1999}{2001}\)
Ta có : \(\frac{1}{3}=\frac{1.2}{3.2}=\frac{2}{6}\); \(\frac{1}{6}=\frac{1.2}{6.2}=\frac{2}{12}\); \(\frac{1}{10}=\frac{1.2}{10.2}=\frac{2}{20}\).Ta được:
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{y.\left(y+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{y.\left(y+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\left[2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{y}-\frac{1}{\left(y+1\right)}\right)\right]=\frac{1999}{2001}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{\left(y+1\right)}\right)=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(y+1\right)}=\frac{1999}{2001}\div2=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{\left(y+1\right)}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Leftrightarrow\left(y+1\right).1=1.2001\)
\(\Leftrightarrow y+1=2001\)
\(\Leftrightarrow y=2001-1=2000\)
Vậy \(x=2000\)
