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

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

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

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

\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)

nâng cao phát triển toán 7 đấy 

mấy bài đấu thì phải

Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)

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

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

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

Hay: \(P=\frac{1}{2012}\)

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

bấm vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm bạn ạ

94 nếu nháy chuột vào đúng 0 sẽ cho biết đúng hay sai đó bạn

Ta có :

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

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

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}\right)}\)

\(\frac{1}{2012}\)

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+

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

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

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

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

\(=\frac{1}{2012}\)

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

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

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

\(=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}\right)\)

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