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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
có :
\(B=\frac{10^{2015}+1}{10^{2014}+1}>1\)
\(\Rightarrow\frac{10^{2015}+1}{10^{2014}+1}>\frac{10^{2015}+1+9}{10^{2014}+1+9}\) \(=\frac{10^{2015}+10}{10^{2014}+10}=\frac{10.\left(10^{2014}+1\right)}{10.\left(10^{2013}+1\right)}\)
\(=\frac{10^{2014}+1}{10^{2013}+1}=A\)
\(\Rightarrow B>A\)
Vậy B > A
k cho mk nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(2004A=\frac{2014^{2015}+2014}{2014^{2015}+1}=\frac{2014^{2015}+1+2013}{2014^{2015}+1}=1+\frac{2013}{2014^{2015}+1}\)
\(2014B=\frac{2014^{2014}+2014}{2014^{2014}+1}=\frac{2014^{2014}+1+2013}{2014^{2014}+1}=1+\frac{2013}{2014^{2014}+1}\)
Vì \(\frac{2013}{2014^{2015}+1}< \frac{2013}{2014^{2014}+1}\)
=> \(1+\frac{2013}{2014^{2015}+1}< 1+\frac{2013}{2014^{2014}+1}\)
=> \(A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gợi ý nhé: bạn hãy so sánh 2014A và 2014B rồi suy ngược lại A và B
Ta có:
2014A=20142014+ 2014/20142014+1=1+2013/20142014+1
2014B=20142013+2014/20142013+1=1+2013/20142013+1
vì 1+2013/20142014+1<1+2013/20142013+1 nên 10A < 10B
suy ra A<B
Ta có: \(A=\frac{2014^{2014}+1}{2014^{2015}+1}\)
\(\Rightarrow2014A=\frac{2014^{2015}+2014}{2014^{2015}+1}=1+\frac{2013}{2014^{2015}+1}\)
\(B=\frac{2014^{2013}+1}{2014^{2014}+1}\)
\(\Rightarrow2014B=\frac{2014^{2014}+2014}{2014^{2014}+1}=1+\frac{2013}{2014^{2014}+1}\)
Mà \(\frac{2013}{2014^{2015}+1}< \frac{2013}{2014^{2014}+1}\Rightarrow1+\frac{2013}{2014^{2015}+1}< \frac{2013}{2014^{2014}+1}\Rightarrow2009A< 2009B\)
\(\Rightarrow A< B\)
Vậy A < B