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

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có chia hết

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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

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+

Ta có: Tử là:

B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\)            (2013 số hạng 1)

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

  =\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)

 =\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)

=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014

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@@@@@

A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}=\frac{1}{2014}\)                                                                                                                                                                                                       

A=\(\frac{1}{2014}\)

xét mẫu ta được 

(2012/2+1)+(2011/3+1)+...+(1/2013+1)

=2014/2+2014/3+...+2014/2013

=2014(1/2+1/3+...+1/2013) (1)

mà tử bằng 1/2+1/3+1/4+..+1/2013 (2)

(1),(2)=> A=1/2014

xét mẫu

2012+2012/2+2011/3+...+1/2013 

=(1+1+1+…+1) + 2012/2+2011/3+...+1/2013 

2012 số hạng

=(1 + 2012/2) + (1 + 2011/3) + ….+ (1+1/2013) 

=2014/2 + 2014/3 + …. + 2014/2013 

=2014 x (1/2 + 1/3 + … + 1/2013) 

=))

(1/2+1/3+1/4+...+1/2013)/(2012+2012/2+2011/3+...+1/2013) =

(1/2+1/3+1/4+...+1/2013)/ 2014 x (1/2+1/3+1/4+...+1/2013) =  1/2014

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