A=2015/2016+2016/2017 và B=2015+2016/2016+2017 .hãy so sánh A và B
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\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
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Đặt 2015.2016+2016=n
suy ra A=(n+1)/n và B=(n+2)/(n+1)
Ta có A - B=(n+1)/n -(n+2)/(n+1)=((n+1)2-n(n+2))/n(n+1)=(n2+2n+1-n2-2n)/n(n+1)=1/n(n+1)
Vì A-B lớn hơn 0 nên A>B
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- \(A=\frac{2015}{2016}+\frac{2016}{2017}>1;\)
- \(B=\frac{2015+2016}{2016+2017}< 1\)
- Nên A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
1 - A = 1 - 2014/2015 = 1/2015
1 - B = 1 - 2015/ 2016 = 1/2016
Vì 1/2015 > 1/2016 => 1 - 2014/2015 > 1 - 2015 / 2016
Hay 1 - A > 1 -B => A < B
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A=2015/2016+2016/2017+2017/2018>2015/2018+2016/2018+2017/2018
=6048/2018>1
B=2015+2016+2017/2016+2017+2018=6048/6051<1
=>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: B = 2015 + 2016 + 2017/2016 + 2017 + 2018
B= 2015 / (2015 + 2016+2017) + 2016/(2016+2017+2018) + 2017/(2016 + 2017 + 2018)
vì 2015/2016 > 2015/(2016 + 2017+2018) ; 2016/2017>2016/(2016+2017+2018) ; 2017/2018 > 2017/(2016+2017+2018)
=> A>B
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\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)
\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)
Suy ra \(A>B\).
có cách nhanh hơn