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PQ
0
G
4 tháng 3 2018
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: \(P=\frac{1}{2012}\)
9 tháng 8 2017
Ta có :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}\right)}\)
\(\frac{1}{2012}\)
55
19 tháng 6 2019
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
19 tháng 6 2019
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}\)
\(=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....\left(\frac{1}{2011}+1\right)+1\)
\(=\frac{2012}{2}+\frac{2012}{3}+\frac{2012}{4}+.....+\frac{2012}{2011}+\frac{2012}{2012}\)
\(=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}\right)\)
Thay vào,rút gọn là ra
10 tháng 11 2015
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
28 tháng 12 2016
Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)
Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)
\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)
\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)
\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)
\(=1012538,5\)
Vậy ....
DP
14 tháng 8 2017
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
16 tháng 12 2015
\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+..+\frac{1}{1+2+3+..+2011}\)
\(S=1+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+...+\frac{1}{2011.\left(2011+1\right):2}\)
\(S=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2011.2012}\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{\cdot3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(S=1+2\left(\frac{1}{2}-\frac{1}{2012}\right)\)
\(S=1+2.\frac{1}{2}-2.\frac{1}{2012}\)
\(S=1+1-\frac{1}{1006}\)
\(S=\frac{2011}{1006}\)
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