cho A=\(\frac{7^{2011}+1}{7^{2013}+1}\)và B=\(\frac{7^{2013}+1}{7^{2015}+1}\)HÃY SO SÁNH A VÀ B
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2N
3
NN
14 tháng 11 2019
\(\frac{A}{B}=\frac{7^{2013}+1}{7^{2014}+1}.\frac{7^{2015}+1}{7^{2014}+1}=\frac{7^{4028}+7^{2013}+7^{2015}+1}{7^{4028}+2.7^{2014}+1}=\)
\(=\frac{7^{4028}+7^{2013}\left(1+7^2\right)+1}{7^{4028}+2.7.7^{2013}+1}=\frac{7^{4028}+50.7^{2013}+1}{7^{4028}+14.7^{2013}+1}>1\)
\(\Rightarrow A>B\)
XO
31 tháng 1 2020
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
XN
0
NH
0
NH
0
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, \(A-B=\frac{-1}{14}\)
L
0
M
11 tháng 8 2018
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
M
11 tháng 8 2018
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)