Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)

\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)

\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)

B=2012.A

=>A/B=1/2012

a/b= 1/2012 nha bạn 

tích

bạn ơi tính A đi bạn

kobiet

Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.

Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)

= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)

= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

= \(1-\frac{1}{2011}\)

= \(\frac{2010}{2011}\)

\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-\frac{2011}{2010}\right)\)

\(=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{1010}\right).\left(1-\frac{3}{2010}\right)....\left(1-\frac{2010}{2010}\right).\left(1-\frac{2011}{2010}\right)\)

\(=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...\left(1-1\right).\left(1-\frac{2011}{2010}\right)\)

\(=\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right).\left(1-\frac{3}{2010}\right)...0.\left(1-\frac{2011}{2010}\right)\)

\(=0\)

*2010/1+2009/2+...+1/2010

=(2009/2+1)+(2008/3+1)+...+(1/2010+1)+1

=2011/2+2011/3+..+2011/2010+2011/2011

=2011(1/2+1/3+1/4+...+1/2011)

=> C=2011/1=2011