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Akai Haruma
Giáo viên
22 tháng 6 2023

Đề có vấn đề. Bạn coi lại.

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

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

4 tháng 7 2016

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

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

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

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

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

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

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

\(x+1=2011\)

\(x=2010\)