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DT
3
20 tháng 7 2016
Đặt tử số là A = 1 + 2 + 22 + 23 + ... + 22012
2A = 2 + 22 + 23 + 24 + ... + 22013
2A - A = (2 + 22 + 23 + 24 + ... + 22013) - (1 + 2 + 22 + 23 + ... + 22012)
A = 22013 - 1
=> \(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
WT
1
21 tháng 4 2019
Đặt A = 1 + 2 + 22 + 23+ ...+ 22012
2A = 2 + 22 + 23 + 24 +....+22013
Lấy 2A - A = 2 + 22 +23 + 24 +....+22013 - 1-2-22- 23 - ... - 22012
A = 22013 - 1
Khi đó : M = A / 22014 -2
= 22013 - 1 / 2.( 22013 - 1 )
= 1/2
Vậy M= 1/2
NT
3
D
8 tháng 3 2016
\(2M=\frac{2+2^2+2^3+...+2^{2013}}{2^{2014}-2}\)
\(2M-M=\frac{\left(2+2^2+...+2^{2013}\right)-\left(1+2^2+...+2^{2012}\right)}{2^{2014}-2}\)
\(M=\frac{1-2^{2013}}{2^{2014}-2}\)
8 tháng 3 2016
Đặt A=1+2+22+............+22012
2A=2+22+23+..............+22013
2A-A=22013-1
A=22013-1
=>M=\(\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2^{2013}.2-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
S
15 tháng 7 2017
Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)
\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)
\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)
\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)
HL
26 tháng 3 2017
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)
11 tháng 7 2016
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
Tính \(A=1+2+2^2+...+2^{2012}\Rightarrow2.A=2.\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow2.A=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow2.A-A=2+2^2+2^3+...+2^{2013}-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(\Rightarrow A=2+2^2+2^3+...+2^{2013}-1-2-2^2-2^3-...-2^{2012}\)
\(\Rightarrow A=2^{2013}-1\)
vậy \(M=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)