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

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\(A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2018^2}\)

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}< \frac{3}{4}\)

Ta có :

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

Vậy 2017³ + 2017² < 2018³

(x+6)⁴=4096

(x+6)⁴=8⁴

==> x+6=8 hoặc x+6=—8

==> x=8–6 hoặc x=—8–6

==> x= 2 hoặc x=—14

2^x—3=128

2^x—3=2⁷

==> x—3=7

x=7+3

x=10

Ss: 2²⁰¹⁸ và 16⁹⁰⁰

Ta có 16⁹⁰⁰=(2⁴)⁹⁰⁰=2³⁶⁰⁰

Vì 2²⁰¹⁸<2³⁶⁰⁰

Nên 2²⁰¹⁸<2³⁶⁰⁰

17²⁰ và 4⁴⁰

Ta có 17²⁰=(17²)¹⁰=289¹⁰

4⁴⁰=(4⁴)¹⁰=256¹⁰

vì 289¹⁰>256¹⁰

Nên 17²⁰ > 4⁴⁰

Cách khác

Ta có 4⁴⁰=4^2.20=(4²)²⁰=16²⁰

Vì 16²⁰< 17²⁰

Nên 17²⁰> 4⁴⁰

333⁴⁴⁴ và 444³³³

Ta có 333⁴⁴⁴=(3.111)^4.111=3^4.111.111^4.111=81¹¹¹.111^4.111

444³³³=(4.111)^3.111=4^3.111.111^3.111=64¹¹¹.111^3.111

Vì 81¹¹¹.111^4.111> 64¹¹¹.111^3.111

Nên 333⁴⁴⁴> 444³³³

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 : \(A=1+2+2^2+...+2^{2017}\)(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