Ta có :

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

Vậy 2017³ + 2017² < 2018³

A=1+2+2²+2³+...+2^{2017 (1)}

2A=2+2²+2³+2⁴+...+2^{2018 (2)}

Lấy (2) - (1) ta có:

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

Mà B=2²⁰¹⁸-1 =>A=B

b) ta có: B=2017²

B=(2016+1).2017=2016.2017+2017

A=2016.2018

A=2016.(2017+1)=2016.2017+2016

Vì 2016<2017=>A<B

mình nhé

a, A = 1+2+2²+...+2²⁰¹⁷

2A=2+2²+2³+...+2²⁰¹⁸

2A-A=A=2²⁰¹⁸-1

=> A  = B

b, A = 2016.2018 =2016.(2017+1)=2016+2017.2016

B=2017²=2017.2017=2017.(2016+1)=2017.2016+2017

Vì 2016 < 2017 => 2016+2017.2016 < 2017.2016+2017 => A < B

cảm ơn nhé, Hoàng

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

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

a) \(A=2+2^2+2^3+...+2^{2017}\)

\(A=2\left(1+2^1+2^2+...+2^{2016}\right)\)

\(A=2.\dfrac{2^{2016+1}-1}{2-1}\)

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

Câu b bạn xem lại đề

a)

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

\(A=3+3^2+3^3+...+3^{2016}\)

\(3A=3^2+3^3+3^4+...+3^{2017}\)

\(2A=3^{2017}-3\)

\(A=\frac{3^{2017}-3}{2}< 3^{2017}-3\)

\(\Rightarrow A< B\)