Mẫu số của A \(=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)

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

      (2012 số 1)                 (2011 phân số)

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

\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)

\(=2013.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)

=> \(A=\frac{1}{2013}\)

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}\)

\(\Rightarrow A=\frac{1}{2013}\)

Vậy \(A=\frac{1}{2013}\)

lol

Xét tử:

\(2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}\)

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

= \(\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)

= \(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)

Thay vào ta có:

A = \(\frac{2013\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}\)

=> A = 2013 

Mà 2013 chia hết cho 3

=> A chia hết cho 3

A = 2013  chia hết cho 3 nhé

http://d.f24.photo.zdn.vn/upload/original/2016/02/14/10/03/3204324726_616688374_574_574.jpg

có chia hết

giải thích ra với

= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)

= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)

= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)

= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)

= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)

A=\(\frac{1+\frac{2011}{2}+1+\frac{2010}{3}+1+...+\frac{1}{2012}+1+1}{\frac{1}{2}+...+\frac{1}{2013}}\)

A=\(\frac{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+...+\frac{1}{2013}}\)

A=\(\frac{2013\left(\frac{1}{2}+...+\frac{1}{2013}\right)}{\frac{1}{2}+...+\frac{1}{2013}}\)

A=2013

Mà 2013: 3 = 671

Vậy A : 3 dư 0 hay\(A⋮3\)

vì sao bạn lại 1+

=1/2014

câu này ở violympic có mà

TA TÁCH 2012 RA THÀNH 2012 CON SỐ 1.LẤY (1 + 2012/2) + (1 + 2011/3) + (1 + 2010/4); +...+ (1 + 1/2013) Ở MẪU, TA ĐƯỢC 2014/2 + 2014/3 +...+ 2014/2013(Ở MẪU).ĐẶT THỪA SỐ CHUNG 2014 RA NGOÀI TA SẼ ĐƯỢC 2014(1/2 + 1/3 +...+ 1/2013)(Ở MẪU).LẤY TỬ CHIA MẪU TA SẼ CÒN LẠI 1/2014. VẬY A=1/2014

Giải:

(1+1/2!)+(1+2/3!)+(1+3/4!)+....+(1+2011/2012!)=2011+(1/2!+2/3!+3/4!+...+2011/2012!)

=2011+(\(\frac{1}{2!}\)+\(\frac{3-1}{3!}\)+\(\frac{4-1}{4!}\)+...+\(\frac{2012-1}{2012!}\))= 2011 +(\(\frac{1}{2!}\)+\(\frac{1}{2!}\)-\(\frac{1}{3!}\)+\(\frac{1}{3!}\)-\(\frac{1}{4!}\)+...+\(\frac{1}{2011!}\)-\(\frac{1}{2012!}\))

= 2011+(1-\(\frac{1}{2012!}\))=2012 - \(\frac{1}{2012!}\)<2012 (đpcm)

cm nha

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