chịu

\(3B=1.3^2+2.3^3+3.3^4+...+2022.3^{2023}+2023.3^{2024}\)

\(2B=3B-B=-3-3^2-3^3-...-3^{2023}+2023.3^{2024}\)

\(2B=2023.3^{2024}-\left(3+3^2+3^3+...+3^{2023}\right)\)

Đặt 

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

\(3C=3^2+3^3+3^4+...+3^{2024}\)

\(2C=3C-C=3^{2024}-3\Rightarrow C=\dfrac{3^{2024}-3}{2}\)

\(\Rightarrow2B=2023.3^{2024}-\dfrac{3^{2024}-3}{2}=\)

\(=\dfrac{2.2023.3^{2024}-3^{2024}+3}{2}=\dfrac{4045.3^{2024}+3}{2}\)

\(\Rightarrow B=\dfrac{4045.3^{2024}+3}{4}\)

\(4A-3^{2023}\) hay \(4A=3^{2023}\) hả bạn

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

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

\(3A-A=\left(1-3+3^2-3^3+...+3^{2021}-3^{2022}\right)-\left(3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\right)\)

\(2A=3^{2023}-1\)

\(\Rightarrow A=\left(3^{2023}-1\right)\div2\)

\(\text{cái này mình sợ sai nên bạn có thể nhờ cô chữa}\)

4A - 3^{2023 nha bạn}

Lời giải:
$A=\frac{1}{4}(1-3+3^2-3^3+...+3^{2022}-3^{2023})$

$3A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024})$

$3A+A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024}+1-3+3^2-3^3+...+3^{2022}-3^{2023})$

$4A=\frac{1}{4}(1-3^{2024})$

$A=\frac{1}{16}(1-3^{2024})$

1+2+2 mũ 2 nhé

Sorry

A = 1+2+2^2+2^3+...+2^2023

=> 2A = 2+2^2+2^3+2^4+...+2^2024

=> 2A-A=A = 2^2024 - 1

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

`5^2023 : 5^2022 + 4^3 - 75`

`= 5^1 + 64 - 75`

`= 69 - 75`

`= -6`

\(5^{2023}:5^{2022}+4^3-75=5+64-75=69-75-6\)

Ta có:

\(2023^{2022}=2023\cdot2023^{2021}\)

\(2022^{2022}+2022^{2021}=2022^{2021}\cdot\left(2022+1\right)=2023\cdot2022^{2021}\)

Mà: \(2023>2022\)

\(\Rightarrow2023^{2021}>2022^{2021}\)

\(\Rightarrow2023^{2021}\cdot2023>2022^{2021}\cdot2023\)

\(\Rightarrow2023^{2022}>2022^{2022}+2022^{2021}\) 

Vậy: ... 

\(A=7^{2024}-7^{2023}+7^{2022}-7^{2021}+...+7^2-7\)

=>\(7A=7^{2025}-7^{2024}+7^{2023}-7^{2022}+...+7^3-7^2\)

=>\(7A+A=7^{2025}-7^{2024}+7^{2023}-7^{2022}+...+7^3-7^2+7^{2024}-7^{2023}+...+7^2-7\)

=>\(8A=7^{2025}-7\)

=>\(A=\dfrac{7^{2025}-7}{8}\)