\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2012^2}-1\right)\)

\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2012^2}\right)\)

\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4048143}{2012\cdot2012}\)

\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)...\left(2011\cdot2013\right)}{\left(2\cdot3\cdot...\cdot2012\right)\left(2\cdot3\cdot...\cdot2012\right)}\)

\(-A=\frac{\left(1\cdot2\cdot...\cdot2011\right)\left(3\cdot4\cdot...\cdot2013\right)}{\left(2\cdot3\cdot...\cdot2012\right)\left(2\cdot3\cdot...\cdot2012\right)}=\frac{1\cdot2013}{2012\cdot2}=\frac{2013}{4024}\)

Cảm ơn bạn Uyên nhiều

ủng hộ mình lên 280 điểm với các bạn

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

\(7832\)

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

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

\(=1+\frac{2011}{2}+\frac{\left(2012+2\right).2011}{2}=1+\frac{2011}{2}+2011.1007\)

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 \(A=\frac{1}{2013}\)

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

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)

Ta có: \(1+2+...+n=\frac{\left(n+1\right)n}{2}\)

\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)

\(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)

\(=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

Vậy nên:

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

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{2011.2014}{2012.2013}\)

\(=\frac{1}{3}.\frac{2014}{2012}=\frac{1007}{3018}\)

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