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

Câu 4:

a) \(\left(-333\right)^{444}=333^{444}=\left[\left(3.111\right)^4\right]^{111}=\left(3^4.111^4\right)^{111}=\left(81.\ 111^4\right)^{111}\)

\(444^{333}=\left[\left(4.111\right)^3\right]^{111}=\left(4^3.111^3\right)^{111}=\left(64.\ 111^3\right)^{111}\)

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 1

a) A = 1 + 2 + 3 + 4 + ...+ 98 + 99 + 100

= (1 + 100) + (2 + 99) +(3+ 98) + (4 + 97)+...+ (50 + 51)

= 101 + 101 + 101+ ... + 101 + 101

= 101 . 50 = 5050

Câu 2:

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

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

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

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

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

x + 2024 = 0

x = - 2024

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

\(101x\ +\ \frac{100+0}{2}\times101\ =\ 8080\)

\(101x\ +\ 5050\ =\ 8080\)

\(101x\ =\ 3030\)

\(x\ =\ 30\)

Câu 3:

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

\(A=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+...+4^{22}\left(1+4+4^2\right)\)

\(A=\left(1+4+4^2\right)\left(4+4^4+...+4^{22}\right)\)

\(A=21.\left(4+4^4+...+4^{22}\right)\)

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

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