SO SÁNH: A=\(\frac{7^{2013}+1}{7^{2014}+1}\) B=\(\frac{7^{2014}+1}{7^{2015}+1}\)
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\(\frac{A}{B}=\frac{7^{2013}+1}{7^{2014}+1}.\frac{7^{2015}+1}{7^{2014}+1}=\frac{7^{4028}+7^{2013}+7^{2015}+1}{7^{4028}+2.7^{2014}+1}=\)
\(=\frac{7^{4028}+7^{2013}\left(1+7^2\right)+1}{7^{4028}+2.7.7^{2013}+1}=\frac{7^{4028}+50.7^{2013}+1}{7^{4028}+14.7^{2013}+1}>1\)
\(\Rightarrow A>B\)
Đặt A= 2015^2013+1/2015^2014+7, B=2015^2014-2/2015^2015-2
2015A= 2015^2014+2015/2015^2014+7= 1 + (2008/2015^2014+7)
2015B= 2015^2015-4030/2015^2015-2= 1 - (4028/2015^2015-2)
Do 2015A>1>2015B nên A>B
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
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, \(A-B=\frac{-1}{14}\)
a = \(\frac{2013}{2014}+\frac{2014}{2015}=\frac{2014-1}{2014}+\frac{2015-1}{2015}\)
\(=1-\frac{1}{2014}+1-\frac{1}{2015}\)
\(=2-\left(\frac{1}{2014}+\frac{1}{2015}\right)>1\) (1)
b = \(\frac{2013+2014}{2014+2015}
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Ta có :
\(\frac{2014^{2015}+1}{2014^{2015}+1}\)\(=1\)
\(\frac{2014^{2014}+1}{2014^{2013}+1}\)\(>1\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
ngoài ra a/b>1 thì a+m/b+m > 1 (m thuộc z, m khác 0) và a,b cậu biết rồi đó
a) số số hạng
(2015-1) : 2+1=1008
tổng dãy số
1008 x (2015 +1) :2 = 1016064
a) 1+3+5+7+...+2015
Day tren co so so hang la:
(2015-1):2+1=1008(so hang)
Tong tren bang: (2015+1).1008:2=1016064
b) \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}.\frac{2}{3}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
=\(\left(2015.2014+2014.2013\right).0\)
= \(0\)