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Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{n\left(n+1\right)}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{n\left(n+1\right)}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{n+1}\right)\)
\(=1-\frac{2}{n+1}\)
\(=\frac{n+1}{n+1}-\frac{2}{n+1}\)
\(=\frac{n-1}{n+1}\)
đặt a=1/3+1/6+1/10+...........+2/n(n+1)
1/2a=1/6+1/12+...........+1/n(n+1)
1/2a=1/2.3+1/3.4+........+1/n(n+1)
1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1
1/2a=1/2-1/n+1
a=(1/2--1/n+1):1/2=2003/2004
1/2-1/n+1=2003/2004.1/2
1/2-1/n+1=2003/4008
1/n+1=1/2-2003/4008
1/n+1=1/4008
suy ra n+1=4008
n=4007
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2010}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4020}\)
\(\Rightarrow x+1=4020\)
=> x = 4020 - 1
=> x = 4019
tung từng vế một thôi
bạn nhác quá éo chịu suy nghĩ
bài này dễ vl
Bài 1:
a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
c, ghi lại đề
d, ghi lại đề
Bài 2:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2008\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
x+1=2009
x=2009-1=2008
Vậy x bằng 2008
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{n\left(n+1\right)}=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{n}-\frac{2}{n+1}\)
Tới đây dễ rồi bạn rút gọn rồi tìm n
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2010}{2011}\)
\(\Leftrightarrow n=4021\).