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\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)
\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)
\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}:2\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}:2=\frac{1}{2001}\Rightarrow x+1=2001\Rightarrow x=2000\)
Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
nên \(x=2002-1=2001\)
Vậy x = 2001
= 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
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)