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Vì 20162016 + 1 < 20162017 + 1
\(\Rightarrow\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2016}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow\)A < B
Ta có :
\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)
\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)
\(\Rightarrow A< B\)
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a, \(S=1+2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)
\(\Rightarrow S=2^{2018}-1\)
Vậy : \(S=2^{2018}-1\)
b, Ta có : \(2^{2018}-1< 2^{2018}=2^2.2^{2016}=4.2^{2016}< 5.2^{2016}\)
Vì : \(2^{2018}-1< 4.2^{2016}< 5.2^{2016}\Rightarrow S< 5.2^{2016}\)
Vậy : \(S< 5.2^{2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015+2016}{4033}\)
\(A=\frac{2015}{2016}+\frac{2016}{2017}\) \(B=\frac{2015}{4033}+\frac{2016}{4033}\)
\(\Rightarrow A>B\)
= nhau vì cùng bằng 1
\(\left(6-5\right)^{2016}=1^{2016}=1\)
\(\left(7-6\right)^{2017}=1^{2017}=1\)
\(=>\left(6-5\right)^{2016}=\left(7-6\right)^{2017}\left(=1\right)\)
Ủng hộ nha
Thanks