a) \(A=2+2^2+2^3+...+2^{2022}\)

\(2A=2.\left(2+2^2+2^3+...+2^{2022}\right)\)

\(2.A=2^2+2^3+2^4+...+2^{2023}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2023}\right)-\left(2+2^2+2^3+...+2^{2022}\right)\)

\(A=2^{2023}-2\)

b) A + 2 = 2^x

Hay \(\left(2^{2023}-2\right)+2=2^x\)

\(2^{2023}-2+2=2^x\)

\(2^{2023}=2^x\)

\(\Rightarrow x=2023\)

   a, A = 2¹ + 2² + 2³ + ...+ 2²⁰²²

     2A =         2² + 2³ +...+ 2²⁰²² + 2²⁰²³

2A - A = 2²⁰²³ - 2¹ 

       A = 2²⁰²³ - 2 

b,   A + 2  = 2\(^x\)  ⇒ 2²⁰²³ - 2  + 2 = 2^\(x\) 

                            2²⁰²³               = 2\(^x\)

                           2023                 = \(x\) 

a) A = 2²⁰²³ - 2

b) x = 2023

\(B=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

\(A=1+3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(1+3+3^2+3^3\right)\left(1+3^4+...+9^{96}\right)\)

\(\Rightarrow A=40\left(1+3^4+...+9^{96}\right)⋮40\)

a) Ta có A = 2¹ + 2² + 2³ + ... + 2²⁰²²

2A = 2² + 2³ + 2⁴ + ... + 2²⁰²³

2A - A = ( 2² + 2³ + 2⁴ + ... + 2²⁰²³ ) - ( 2¹ + 2² + 2³ + ... + 2²⁰²² )

A = 2²⁰²³ - 2

Lại có B = 5 + 5² + 5³ + ... + 5²⁰²²

5B = 5² + 5³ + 5⁴ + ... + 5²⁰²³

5B - B = ( 5² + 5³ + 5⁴ + ... + 5²⁰²³ ) - ( 5 + 5² + 5³ + ... + 5²⁰²² )

4B = 5²⁰²³ - 5

B = \(\dfrac{5^{2023}-5}{4}\)

b) Ta có : A + 2 = 2^x

⇒ 2²⁰²³ - 2 + 2 = 2^x

⇒ 2²⁰²³ = 2^x

Vậy x = 2023

Lại có : 4B + 5 = 5^x

⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5^x

⇒ 5²⁰²³ - 5 + 5 = 5^x

⇒ 5²⁰²³ = 5^x

Vậy x = 2023

1)

Ta có :

2³⁰⁰ = ( 2³ )¹⁰⁰ = 8¹⁰⁰

3²⁰⁰ = ( 3² )¹⁰⁰ = 9¹⁰⁰

vì 8¹⁰⁰ < 9¹⁰⁰ nên 2³⁰⁰ < 3²⁰⁰

2)

Ta có :

5²³ = 5²² . 5

vì 5²² . 5 < 5²² . 6 nên 5²³ < 6 . 5²²

1) \(2^{300}vs3^{200}\) 

Ta có: \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

          \(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì 8 < 9 nên 8¹⁰⁰<9¹⁰⁰ => 2³⁰⁰ < 3²⁰⁰

2) \(5^{23}vs6.5^{22}\)

Ta có: \(5^{23}=5^{22}.5\)

Vì 5²² = 5²² và 5 < 6 nên 5²². 5 < 5²² . 6 => 5²³ < 6.5²²

1 ta có

2³⁰⁰= (2³)¹⁰⁰ = 8¹⁰⁰

3²⁰⁰= (3²)¹⁰⁰ = 9¹⁰⁰

Vì 8¹⁰⁰ < 9¹⁰⁰ vậy 2³⁰⁰ < 3²⁰⁰

2 ta có: 5²³= 5²².5.

vì 5²².5 < 6.5²² nên 5²³ < 6.5²²

\(A=1+2+2^2+2^3+....+2^{98}+2^{99}\\ \Leftrightarrow A=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+....+\left(2^{98}+2^{99}\right)\\ \Leftrightarrow A=3+2^2.\left(1+2\right)+2^4.\left(1+2\right)+....+2^{98}.\left(1+2\right)\\ \Leftrightarrow A=3+3.2^2+3.2^4+....+3.2^{98}\\ \Leftrightarrow A=3.\left(1+2^2+2^4+...+2^{98}\right)⋮3\)

Đặt A=1 + 2 + 2²+ 2³+ 2⁴ +... + 2⁹⁹ + 2¹⁰⁰

=>2A=2 + 2²+ 2³+ 2⁴ +... + 2⁹⁹ + 2¹⁰⁰+2¹⁰¹

=>2A-A=(2 + 2²+ 2³+ 2⁴ +... + 2⁹⁹ + 2¹⁰⁰+2¹⁰¹)-(1 + 2 + 2²+ 2³+ 2⁴ +... + 2⁹⁹ + 2¹⁰⁰)

=>A=2¹⁰¹-1

41+42+43+44-21-22-23-24

=( 41- 21 )+ (42-22)+(43-23)+(44-24)

=20 + 20 +20 +20

=20 . 4

=80

41+ 42 + 43 + 44 - 21 - 22 - 23 - 24 = 80

41 + 42 + 43 + 44 - 21 - 22 - 23 - 24

= ( 41 - 21 ) + ( 42 - 22 ) + ( 43 - 23 ) + ( 44 - 24 )

= 20 + 20 + 20 + 20

= 80

\(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+2^5+...+2^{100}+2^{101}\)

\(\Rightarrow2A-A=2^{101}-1\)

\(\Leftrightarrow A=2^{101}-1\)

Đặt biểu thức là A

ta có 2A-A=2^101-1

Đặt \(A=1+2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)=2^{101}-1\)