HOC24
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{x-1}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{2001}:2\)
\(\frac{1}{2}-\frac{1}{x-1}=\frac{1999}{4002}\)
\(\frac{1}{x-1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{x-1}=\frac{1}{2001}\)
\(\Rightarrow x+1=\frac{1.2001}{1}=2001\)
\(\Rightarrow x=2001-1\)
\(\Rightarrow x=2000\)
Vậy \(x=2000\)
=> \(-\frac{2}{5}+\frac{7}{5}\le x