Bài 1:

$B=1+3+3^2+3^3+...+3^{100}$

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

$=1+3(1+3)+3^3(1+3)+...+3^{99}(1+3)$

$=1+(1+3)(3+3^3+...+3^{99})=1+4(3+3^3+....+3^{99})$

$\Rightarrow B$ chia 4 dư 1.

Bài 2:

$C=5-5^2+5^3-5^4+...+5^{2023}-5^{2024}$

$5C=5^2-5^3+5^4-5^5+...+5^{2024}-5^{2025}$

$\Rightarrow C+5C=5-5^{2025}$

$6C=5-5^{2025}$

$C=\frac{5-5^{2025}}{6}$

a: \(24\cdot35+65\cdot24-1400\)

\(=24\left(35+65\right)-1400\)

\(=24\cdot100-1400=2400-1400=1000\)

b: \(32:2^4-4\cdot5^2\)

\(=\dfrac{32}{16}-4\cdot25\)

=2-100

=-98

c: \(17+75:\left[30-5\cdot\left(7^2-48\right)\right]\)

\(=17+75:\left[30-5\cdot\left(49-48\right)\right]\)

\(=17+75:\left[30-5\right]\)

\(=17+\dfrac{75}{25}=17+3=20\)

d: \(\left(-138\right)-\left[258+4\cdot\left(-36\right)\right]\cdot2024^0\)

\(=-138-\left[258-144\right]\)

\(=-138-114\)

=-252

300

300

= 300 

a: \(61\cdot45+61\cdot23-68\cdot51\)

\(=61\left(45+23\right)-68\cdot51\)

\(=68\cdot61-68\cdot51\)

\(=68\left(61-51\right)=68\cdot10=680\)

b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)

\(=75-75+4\cdot8\)

\(=4\cdot8=32\)

c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)

\(=\dfrac{36}{20-30+4^2}\)

\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)

d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)

\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)

A = 5 + 5² + 5³ + ... + 5²⁰²³

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)

= 5²⁰²⁴ - 5

⇒ A = (5²⁰²⁴ - 5)/4

A = 5 + 5² + 5³ + ... + 5²⁰²³

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)

= 5²⁰²⁴ - 5

⇒ A = (5²⁰²⁴ - 5)/4

a) \(S=1+2+2^2+..+2^{2022}\)

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

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

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

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

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

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

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

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

2² x 4³ - 625 : {[504 - (5² x 8 + 70) : 3² + 6] : 20 + 2023⁰}

= 4 x 64 - 625 : {[504 - (25 x 8 + 70) : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - (200 + 70) : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - 270 : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - 30 + 6] : 20 + 1}

= 256 - 625 : {[474 + 6] : 20 + 1]

= 256 - 625 : {480 : 20 + 1]

= 256 - 625 : {24 + 1}

= 256 - 625 : 25

= 256 - 41

= 215

22 x 43 - 625 : {[504 - (52 x 8 + 70) : 32 + 6] : 20 + 20230}

= 4 x 64 - 625 : {[504 - (25 x 8 + 70) : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - (200 + 70) : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - 270 : 9 + 6] : 20 + 1}

= 256 - 625 : {[504 - 30 + 6] : 20 + 1}

= 256 - 625 : {[474 + 6] : 20 + 1]

= 256 - 625 : {480 : 20 + 1]

= 256 - 625 : {24 + 1}

= 256 - 625 : 25

= 256 - 41

= 215

625:25=25 mà

Ta có : 
A = 1 + 5 + \(5^2\)+\(5^3\)+...+ \(5^{2023}\)
5A = 5 + \(5^2\)+\(5^3\)+\(5^4\)+..+ \(5^{2024}\)
=> 5A - A = ( 5 + \(5^2\)+\(5^3\)+\(5^4\)+..+ \(5^{2024}\) ) - ( 1 + 5 + \(5^2\)+\(5^3\)+...+ \(5^{2023}\) ) 
=> 4A =  \(5^{2024}\)- 1
Nhận thấy : 
                  \(5^{2024}\) - 1 > ​​​\(5^{2024}\)
=> 4A <  \(5^{2024}\) 
                            Vậy 4A <  \(5^{2024}\) ​

Thấy hay tick hộ mk vs ạ