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6 tháng 4 2016

Nhân cả 2 vế với 1/2 ta có:

1/6+ 1/12+ 1/20+... +1/x(x+1)=2009/4022

1/2.3+ 1/3.4+ 1/4.5+... +1/x(x+1)=2009/4022

1/2- 1/3+ 1/3-1/4+ 1/4- 1/5+...+1/x- 1/x+1=2009/4022

1/2- 1/x+1=2009/4022

x+1/2.(x+1)- 2/2.(x+1)=2009/4022

x+1-2/2.(x+1)=2009/4022

x-1/2.(x+1)=2009/4022

=>(x-1). 4022= 2009.2.(x+1)

 4022.x- 4022= 4018.x+ 4018

 4022.x-4018.x= 4018+ 4022

 x.(4022- 4018)= 8040

 x.4                = 8040

 x                   = 8040: 4

 x                   =2010

AH
Akai Haruma
Giáo viên
22 tháng 6 2023

Đề có vấn đề. Bạn coi lại.

4 tháng 7 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2011}\)

\(x+1=2011\)

\(x=2010\)

16 tháng 4 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\Rightarrow x+1=2011\Rightarrow x=2010\)

Vậy x=2010

28 tháng 10 2019

hộ mk nha bạn nhanh 1h mk cần r

28 tháng 10 2019

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)

\(x+1=2011\)

\(x=2010\)

12 tháng 8 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(=>\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)

\(=>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1009}{4022}\)

\(=>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(=>\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

\(=>\frac{1}{x+1}=\frac{1}{2011}\)

\(=>x+1=2011\)

\(=>x=2010\)

12 tháng 8 2016

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)

\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x.\left(x+1\right):2}\right):2=\left(\frac{2009}{2011}\right):2\)

\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

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

=> x + 1 = 2011

=> x = 2000