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1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1

2. A>B

mik nghĩ là = các bạn k mi nha 

ai giải hộ mình  với

câu trả lời là ; <

giải chi tiết giúp tớ đi

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 :

\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)

\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)

\(\Rightarrow A< B\)

A<B

nha bạn

A<B

nhá bạn

thanks :)

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 : \(S< 5.2^{2016}\)

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 !!! 

\(A=\frac{2015}{2016}+\frac{2016}{2017}\)                                               \(B=\frac{2015+2016}{4033}\)

\(A=\frac{2015}{2016}+\frac{2016}{2017}\)                                           \(B=\frac{2015}{4033}+\frac{2016}{4033}\)

\(\Rightarrow A>B\)