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\(2015.2017-\frac{1}{2014}+2015.2016\)
\(=2015.\left(2017+2016\right)-\frac{1}{2014}\)
\(=2015.4033-\frac{1}{2014}\)
\(=8126495\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2015\cdot2017-1}{2014+2015\cdot2016}\)\(\cdot\frac{2}{3}\)
\(=\frac{2015\cdot\left(2016+1\right)-1}{2014+2015\cdot2016}\cdot\frac{2}{3}\)
\(=\frac{2015\cdot2016+\left(2015-1\right)}{2014+2015\cdot2016}\cdot\frac{2}{3}\)
\(=\frac{2015\cdot2016+2014}{2014+2015\cdot2016}\cdot\frac{2}{3}\)
\(=1\cdot\frac{2}{3}\)
\(=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\frac{2}{2015+2016}>\frac{2}{2016+2017}\rightarrow2-\frac{2}{2015+2016}< 2-\frac{2}{2016+2017}\)
\(\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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S = 2020 + 2019 - 2018 - 2017 + 2016 + 2015 - 2014 - 2013 + ... + 4 + 3 - 2 - 1
= ( 2020 + 2019 - 2018 - 2017 ) + ( 2016 + 2015 - 2014 - 2013 ) + ... + ( 4 + 3 - 2 - 1 ) (có tất cả 2020 : 4 = 505 nhóm)
= 4 + 4 + ... + 4
= 4. 505 = 2020
Vậy S = 2020.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=2015\left(2017+2016\right)-\frac{1}{2014}\)
\(=2015.4033-\frac{1}{2014}\)
\(=8126495-\frac{1}{2014}=8126495\)