so sanh a= 2015^2014+1/2015^2014-1 va b= 2015^2014-1/2015^2014-3
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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 \(A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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CÁCH 1:
A=1và 2/2015^2014-1
B= 1và 2/2015^2014-3
Vì 1và 2/2015^2014-1 < 1và 2/2015^2014-3
Vậy A <B
CÁCH 2:
Ta biết: a/b>1=>a/b> a+n/b+n
B>1=> B= 2015^2014-1/2015^2014-3> 2015^2014-1+2/2015^2014-3+2=2015^2014+1/2015^2014-1=A
Vậy B>A
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 99^2015 + 1/99^2014 + 1 < 99^2015 + 1 + 98 / 99^2014 + 1 + 98
= 99^2015 + 99 / 99^2014 + 99
= 99(99^2014 + 1) / 99(99^2013+1)
= 99^2014 + 1 / 99^2013 + 1 = B
=> A < B
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Giải:
Ta có:
\(A=\frac{2014+2015}{2015+2016}=\frac{2014+2015+2}{2015+2016}-\frac{2}{2015+2016}=2-\frac{2}{2015+2016}\)(1)
\(B=\frac{2015+2016}{2016+2017}=\frac{2015+2016+2}{2016+2017}-\frac{2}{2016+2017}=2-\frac{2}{2016+2017}\)(2)
Từ (1) và (2) ta có: \(A=2-\frac{2}{2015+2016}\)và \(B=2-\frac{2}{2016+2017}\)
Vì \(\frac{2}{2015+2016}>\frac{2}{2016+2017}\rightarrow2-\frac{2}{2015+2016}< 2-\frac{2}{2016+2017}\)
\(\Rightarrow A< B\)
\(A=\frac{2015^{2014}+1}{2015^{2014}-1}=\frac{2015^{2014}-1+2}{2015^{2014}-1}=1+\frac{2}{2015^{2014}-1}.\)
\(B=\frac{2015^{2014}-1}{2015^{2014}-3}=\frac{2015^{2014}-3+2}{2015^{2014}-3}=1+\frac{2}{2015^{2014}-3}\)
mà \(\frac{2}{2015^{2014}-1}< \frac{2}{2015^{2014}-3}\)( 20152014 -1 > 20152014 - 3)
\(\Rightarrow A< B\)