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![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)
\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)
\(Vậy:A>B\)
Đúng nha Nguyễn Bình Minh
so sánh:
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\) và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)
\(B=\frac{2013}{2014+2015+2016}+\frac{2014}{2014+2015+2016}+\frac{2015}{2014+2015+2016}\)
Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)
\(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)
\(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}>\frac{2013+2014+2015}{2014+2015+2016}\)
Vậy: \(A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1-1/(2013*2014)
B=1-1/(2014*2015)
2013*2014<2014*2015
=>1/2013*2014>1/2014*2015
=>-1/2013*2014<-1/2014*2015
=>A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2013}{2013}+\frac{1}{2013}+\frac{1}{2013}=\left(\frac{2013}{2014}+\frac{1}{2013}\right)+\left(\frac{2014}{2015}+\frac{1}{2013}\right)+1\)
Ta có: \(\frac{2013}{2014}+\frac{1}{2013}>\frac{2013}{2014}+\frac{1}{2014}=\frac{2014}{2014}=1\)
\(\frac{2014}{2015}+\frac{1}{2013}>\frac{2014}{2015}+\frac{1}{2015}=\frac{2015}{2015}=1\)
=> A > 1+ 1 + 1 = 3
![](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)
Ta có :
\(\frac{2013}{2014}< 1\)( 1 )
\(\frac{2014}{2015}< 1\)( 2 )
\(\frac{2015}{2016}< 1\)( 3 )
từ ( 1 ) , ( 2 ) và ( 3 )
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}< 1+1+1=3\)
vậy A < 3
Có: 2013/2014<2014/2014
2014/2015<2015/2015
2015/2016<2016/2016
=>2013/2014+2014/2015+2015/2016<2014/2014+2015/015+2016/2016
=>A<3
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\dfrac{2013+2014}{2014+2015}=\dfrac{2013}{2014+2015}+\dfrac{2014}{2014+2015}\)
B=\(\dfrac{2013}{2014}+\dfrac{2014}{2015}\)
Vì \(\dfrac{2013}{2014}>\dfrac{2013}{2014+2015}\); \(\dfrac{2014}{2015}>\dfrac{2014}{2014+2015}\) nên B>A