Thực hiện phép tính: A = 1 + 2 + 3 + 4 + ...+ 98 + 99 + 100

Câu 4:

a) ${(-333)}^{444}={333}^{444}={\left[{\left(3.111\right)}^4\right]}^{111}={\left(3^4{.111}^4\right)}^{111}={\left(81.{111}^4\right)}^{111}$

${444}^{333}={\left[{\left(4.111\right)}^3\right]}^{111}={\left(4^3{.111}^3\right)}^{111}={\left(64.{111}^3\right)}^{111}$

Ta thấy 64 < 81

3 < 4 nên 111³ < 111⁴

=> (81. 111⁴)¹¹¹ > (64. 111³)¹¹¹

Vậy (- 333)⁴⁴⁴ > 444³³³

b) $1-\frac{2020}{2021}=\frac{1}{2021}$

$1-\frac{2021}{2022}=\frac{1}{2022}$

Ta có: $\frac{1}{2021}>\frac{1}{2022}$ nên $\frac{2020}{2021}<\frac{2021}{2022}$

Câu 2:

a) $\frac{x+4}{2020}+\frac{x+3}{2021}=\frac{x+2}{2022}+\frac{x+1}{2023}$

$\frac{x+4}{2020}+1+\frac{x+3}{2021}+1=\frac{x+2}{2022}+1+\frac{x+1}{2023}+1$

$\frac{x+4+2020}{2020}+\frac{x+3+2021}{2021}=\frac{x+2+2022}{2022}+\frac{x+1+2023}{2023}$

$\frac{x+2024}{2020}+\frac{x+2024}{2021}-\frac{x+2024}{2022}-\frac{x+2024}{2023}=0$

$(x+2024)(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023})=0$

x + 2024 = 0

x = - 2024

b) $x+(x+1)+(x+2)+(x+3)+....+(x+98)+(x+99)+(x+100)=8080$

$101x+\frac{100+0}{2}\times 101=8080$

$101x+5050=8080$

$101x=3030$

$x=30$

Câu 3:

a) $A=4+4^2+4^3+4^4+...+4^{23}+4^{24}$

$A=4(1+4+4^2)+4^4(1+4+4^2)+...+4^{22}(1+4+4^2)$

$A=(1+4+4^2)(4+4^4+...+4^{22})$

$A=21.(4+4^4+...+4^{22})$

Ta có: 21⋮21 => 21.(4 + 4⁴ + ...+ 4²²) ⋮ 21 => A⋮21

b) B = n(n + 1)(n +2)(n +3)

+) Nếu n chẵn thì n⋮2 nên n(n + 1)(n + 2)(n + 3) ⋮ 2 và 4

=> n(n + 1)(n + 2)(n + 3) ⋮ 8

+) Nếu n lẻ thì (n + 1)⋮2 và (n+3)⋮4

=> n(n + 1)(n + 2)(n + 3) ⋮ 8

+) Nếu n = 3 thì => n(n + 1)(n + 2)(n + 3) ⋮ 3

Nếu n chia 3 dư 1 thì n có dạng 3k + 1

Khi đó n + 2 = 3k + 3 = 3(k+1)⋮3

Nếu n chia 3 dư 2 thì n có dạng 3k+2

khi đó n+1 = 3k + 3 = 3(k + 1)⋮ 3

=> n(n + 1)(n + 2)(n + 3) ⋮ 3

Vậy => n(n + 1)(n + 2)(n + 3) ⋮ 24 mọi n nguyên