a)  5²³ và 6*5²²

5²³ = 5 * 5²²

Vì 5<6 suy ra 5 * 5²² < 6 * 5²² hay 5²³ < 6*5²²

Vậy: 5²³ < 6 * 5²²

b) 7 * 2¹³ và 2¹⁶

2¹⁶ = 2³ * 2¹³ = 8 * 2¹³

Vì 7 < 8 suy ra 7 * 2¹³ < 8 * 2¹³ hay 7 * 2¹³ < 2¹⁶

Vậy: 7 * 2¹³ < 2¹⁶

c) 21¹⁵ và 27⁵ * 49⁸

27⁵ * 49⁸ = [(3)³]⁵ * [(7)²]⁸ = 3¹⁵ * 7¹⁶ = 3¹⁵ * 7¹⁵ *7 = (3*7)¹⁵ *7 = 21¹⁵ * 7

Vì 21¹⁵ < 21¹⁵ * 7 suy ra 21¹⁵  < 27⁵ * 49⁸

Vậy: 21¹⁵  < 27⁵ * 49⁸

Ta có:

\(A=1+2+2^2+...+2^{2002}\)

\(2A=2+2^2+2^3+...+2^{2003}\)

\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+....+2^{2002}\right)\)

\(A=2^{2003}-1\)

Mà: \(2^{2003}=2^{2003}\)

\(\Rightarrow2^{2003}-1< 2^{2003}\)

\(\Rightarrow A< B\)

\(A=2+2^2+2^3+\dots+2^{60}\\2A=2^2+2^3+2^4+\dots+2^{61}\\2A-A=(2^2+2^3+2^3+\dots+2^{61})-(2+2^2+2^3+\dots+2^{60})\\A=2^{61}-2\)

Ta thấy: \(2^{61}-2< 2^{61}\)

\(\Rightarrow A< B\)

A=2+2²+2³+...+2⁶⁰

\(\Rightarrow\)2A=2²+2³+2⁴+...+2⁶¹

\(\Rightarrow\)2A-A=(2²+2³+2⁴+...+2⁶¹)-(2+2²+2³2⁴+...+2⁶⁰)

\(\Rightarrow\)A=2⁶¹-2

Mà 2⁶¹-2<2⁶¹ nên A<B

Vậy A<B

\(a< b\)

Có 5mũ 23= 5mũ22 ×5 Mà 5mũ22 =5 mũ22 , 6>5 => 5mũ 23< 6×5 mũ 22

So sánh 5²³ và 6 x 5²²

Ta có:

5²³ = 5 x 5²² 

\(\Rightarrow\)5²³ < 6 x 5²²

a) 123ab chia hết cho 2 và 5 nên b=0

123a0 chia hết cho 9 nên (1+2+3+a+0) chia hết cho 3

=>(6+a) chia hết cho 3

=>a=0;a=3;a=6;a=9

b)3ab chia hết cho 5 nên b=0 hoặc b=5

+với b=5:

3a5 chia 9 dư 4 nên (3+a+5) chia 9 dư 4

=>(8+a) chia 9 dư 4;

=>a=5

mà 355 chia 7 dư 5=>ko thỏa mãn

+với b=0:

3a0  chia 9 dư 4 nên (3+a+0) chia 9 dư 4;

=>(3+a) chia 9 dư 4;

=>a=1

Mà 310 chia 7 dư 2 => số cần tìm là 310

Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)

\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)

\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)

Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)

2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)

2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)

Suy ra  A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2

Vậy A < 2

\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)

\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)

 

 

Sửa:

$2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{20 23}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2$

a) 2⁴ và 4².Ta có:                                                   b)3¹⁶ và 27⁵.Ta có:

2⁴=(2²)²=4²                                                                             27⁵=(3³)⁵=3¹⁵<3¹⁶

=>2⁴=4².Vậy..                                                         =>27⁵<3¹⁶.Vậy...

c)2³³ và 3²².Ta có:                                                 d)chịu

2³³=(2³)¹¹=8¹¹                                                                  

3²²=(3²)¹¹=9¹¹>8¹¹.                                               

=>2³³<3²².Vậy....

a) \(2^4\)

\(4^2=\left(2^2\right)^2=2^4\)

\(\Rightarrow2^4=4^2\)

b) \(3^{16}=3^{16}\)

\(27^5=\left(3^3\right)^5=3^{15}\)

\(\Rightarrow3^{16}>27^5\)

còn thiếu