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AH
Akai Haruma
Giáo viên
22 tháng 6 2023

Đề có vấn đề. Bạn coi lại.

1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009

=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009

=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2

=>1/2-1/(x+1)=2007/4018

=>1/(x+1)=1/2-2007/4018

=>1/x+1=1/2009

=>x+1=2009

=>x=2009-2008

=>x=1

vậy x=1

 

28 tháng 4 2018

làm đúng rồi nhưng phần: 

x+1=2009

x=2009-1

x=2008

mà bạn

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

4 tháng 7 2016

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

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

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

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

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

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

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

\(x+1=2011\)

\(x=2010\)

14 tháng 5 2016

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

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

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

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

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

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

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

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

=>x+1=2011

=>x=2010