Ta có : \(A=1+2+2^2+...+2^{2017}\)(1)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2018}\)(2)

Lấy (2) trừ (1) ta có : 

\(\Rightarrow A=2^{2018}-1\)

\(\Rightarrow A< B\). Vì \(B=2^{2018}\)

A = 1+2+2²+2³+.....+2²⁰¹⁷

2A = 2(1+2+2²+2³+.....+2²⁰¹⁷)  = 2+2²+2³+2⁴+.....+2²⁰¹⁸

2A - A = 2+2²+2³+2⁴+.....+2²⁰¹⁸- (1+2+2²+2³+.....+2²⁰¹⁷)

=> A = 2+2²+2³+2⁴+.....+2²⁰¹⁸-1-2-2²-2³-.....-2²⁰¹⁷

       A =2²⁰¹⁸-1 < 2²⁰¹⁸

Vậy A < B

\(A=1+2+2^2+...+2^{2017}\)

\(\Rightarrow A=\dfrac{2^{2017+1}-1}{2-1}\)

\(\Rightarrow A=2^{2018}-1\)

mà \(B=2^{2018}\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}\)

\(\Rightarrow A-B=-1\)

\(2A=2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow A=2A-A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

Ta có :

\(2A=2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)

\(\Rightarrow A=2^{2018}-1< 2^{2018}=B\)

Vậy A<B

Ta có: \(A=\frac{2^{2017}+2}{2^{2017}+3}=1-\frac{1}{2^{2017}+3}\)

        \(B=\frac{2^{2017}+1}{2^{2017}+2}=1-\frac{1}{2^{2017}+2}\)

Vì \(\frac{1}{2^{2017}+3}< \frac{1}{2^{2017}+2}\) nên \(1-\frac{1}{2^{2017}+3}>1-\frac{1}{2^{2017}+2}\)

hay A > B

A>B bạn nhé

Ta có: \(A=1+2+2^2+...+2^{2017}\)

\(2.A=2+2^2+2^3+...+2^{2018}\)

\(2A-A=2+2^2+2^3+...+2^{2018}-\left(1+2+2^2+...+2^{2017}\right)\)

\(A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

\(A=\frac{2017^{2018+1}}{2017^{2018-3}}\)và \(B=\frac{2017^{2018-1}}{2017^{2018-5}}\)

Có \(A=\frac{2017^{2019}}{2017^{2015}}\)và \(B=\frac{2017^{2017}}{2017^{2013}}\)

Mà\(\frac{2017^{2019}}{2017^{2015}}>\frac{2017^{2018}}{2017^{2015}}\)và\(\frac{2017^{2017}}{2017^{2013}}>\frac{2017^{2017}}{2017^{2015}}\)

Vì \(\frac{2017^{2018}}{2017^{2015}}>\frac{2017^{2017}}{2017^{2015}}\)

Vậy A>B

Ta có :

2017³ + 2017² = 2017² . 2017 + 2017² . 1 = 2017² . ( 2017 + 1 ) = 2017² . 2018 < 2018² . 2018 = 2018³

Vậy 2017³ + 2017² < 2018³