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20 tháng 4 2016

=>2/6+2/12+2/20+...+2/x(x+1)=2013/2015

=>2(1/2.3+1/3.4+1/4.5+...+1/x(x+1)=2013/2015

=>2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)=2013/2015

=>(1/2-1/x+1)=2013/2015:2

=>-(1/x+1)=2013/4030-1/2

=>-(1/x+1)=-(1/2015)=>x+1=2015=>x=2014

2 tháng 4 2015

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)

\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)

\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)

 

2 tháng 4 2015

bạn trên trả lời đúng rùi đó

28 tháng 8 2015

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}:2\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}:2=\frac{1}{2001}\Rightarrow x+1=2001\Rightarrow x=2000\)

17 tháng 2 2018

000000000000000000000000000

17 tháng 5 2016

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

20 tháng 5 2016

= 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

20 tháng 5 2016

x=2002

20 tháng 4 2016

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 \(1:\left(x+1\right)=-\frac{2013}{4030}\)

       \(x+1=-\frac{4030}{2013}\)

\(=>x=-\frac{6043}{2013}\)