1/3 + 1/6 +1/10 +...+ 1/x.(x+1):2 = 2009/2011
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LN
1
TT
3
28 tháng 10 2019
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(x+1=2011\)
\(x=2010\)
NK
2
AL
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
H
0
NM
0
NK
1
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
TN
1
4 tháng 7 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{x+1}=\frac{1}{2011}\)
\(x+1=2011\)
\(x=2010\)
HT
0