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28 tháng 2 2016

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)

=> \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{2002}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)

=> \(\frac{1}{x+1}=\frac{1}{2002}\)

=> x + 1 = 2002

=> x = 2001

8 tháng 4 2018

Ta có: 1/3+1/6+1/10+...+2/x*(x+1)

=2/6+2/12+2/20+...+2/x*(x+1)

=2/2*3+2/3*4+2/4*5+...+2/x*(x+1)

=2*(1/2*3+1/3*4+1/4*5+...+1/x*(x+1))

=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)

=2*(1/2-1/x+1)=2000/2002

=>1/2-1/x+1=2000/2002:2

=>1/2-1/x+1=500/1001

=>1/x+1=1/2-500/1001

=>1/x+1=1/2002

=>x+1=2002

=>x=2002-1

=>x=2001 thuộc N

Vậy x=2001

*Mình ko biết ấn dấu phân số với dấu nhân ở đâu, bạn thông cảm nhé!

8 tháng 4 2018

uk mình cảm ơn bạn rất nhiều 

7 tháng 7 2020

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)=\(\frac{2000}{2002}\)

2.(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2000}{2002}\)

2.\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)

2.(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)) = \(\frac{2000}{2002}\)

2.\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)

\(\frac{1}{x+1}=\frac{1}{2002}\)

2002.1 = (x+1).1

2002 = x+1

x=2001 (T/M)

7 tháng 7 2020

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(\Rightarrow\) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(\frac{1}{2}-\frac{1}{x+1}=\frac{500}{1001}\)

\(\Rightarrow\) \(\frac{1}{x+1}=\frac{1}{2002}\)

\(\Rightarrow\) \(x+1=2002\) \(\Rightarrow\) \(x=2001\)

3 tháng 3 2017

1) Ta có: A=\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\)

=\(\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\)

=\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{1}{3}.\frac{x+2}{x+3}=\frac{125}{376}\)

<=> \(\frac{x+2}{x+3}=\frac{375}{376}\)<=> 376(x+2)=375(x+3) <=> 376x+752=375x+1125 => X=373

x=373 nha bạn

mk đang âm điểm,tk mk nha 

3 tháng 5 2015

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

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

\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)

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

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

Theo bài ra ta có:

\(\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2011}{2013}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}.\frac{1}{2}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2013}{4026}-\frac{2011}{4026}\)

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

=> x + 1 = 2013 

=> x = 2013 - 1 

=> x = 2012 \(\in\) N

Vậy x = 2012

 

 

 

3 tháng 5 2015

Đặt S=1/3+1/6+1/10+..........+2/x(x+1)

1/2S=1/2[1/3+1/6+1/10+...+2/x(x+1)]

1/2S=1/6+1/12+1/20+......1/x(x+1)

1/2S=1/2.3+1/3.4+1/4.5+.....+1x(x+1)

1/2S=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1

1/2S=1/2-1/x+1

Vì S=2011/2013

suy ra (1/2-1/x+1):1/2=2011/2013

(1/2-1/x+1).2=2011/2013

1/2-1/x+1=2011/2013:2

1/2-1/x+1=2011/4026

1/x+1=1/2-2011/4026

1/x+1=1/2013

suy ra x+1=2013

x=2013-1

x=2012