Bài 1:

a: \(S=1-5+5^2-5^3+...+5^{98}-5^{99}\)

=>\(5S=5-5^2+5^3-5^4+...+5^{99}-5^{100}\)

=>\(6S=5-5^2+5^3-5^4+...+5^{99}-5^{100}+1-5+5^2-5^3+...+5^{98}-5^{99}\)

=>\(6S=-5^{100}+1\)

=>\(S=\dfrac{-5^{100}+1}{6}\)

b: S=1-5+5²-5³+...+5⁹⁸-5⁹⁹ là số nguyên

=>\(\dfrac{-5^{100}+1}{6}\in Z\)

=>\(-5^{100}+1⋮6\)

=>\(5^{100}-1⋮6\)

=>\(5^{100}\) chia 6 dư 1

0\(a.S=1-5+5^2-5^3+...+5^{98}-5^{99}\\ 5S=5-5^2+5^3-5^4+.....+5^{99}-5^{100}\\ 5S+S=\left(5-5^2+5^3-5^4+.....+5^{99}-5^{100}\right)+\left(1-5^{ }+5^2-5^3+.....+5^{98}-5^{99}\right)\\ 6S=1-5^{100}\\ S=\dfrac{1-5^{100}}{6}\\ \)

\(b,S6=1-5^{100}\\ 1-S6=5^{100}\) 

=> 5¹⁰⁰ chia 6 du 1

e đang cần gấp, có ai đến giúp e ko?

Đề bài thiếu yêu cầu cụ thể em nhé. em cập nhật lại câu hỏi để được sự hỗ trợ tốt nhất cho tài khoản olm vip

#@₫!%&@^@₫@₫=_++_×%@%@&@@@@=@

Bài 1:

   D     =      5  + 5² + 5³+...+ 5¹⁰⁰

5.D     =             5² + 5³+...+5 ¹⁰⁰ + 5¹⁰¹

5D - D = 5¹⁰¹ - 5

4D       = 5¹⁰¹ - 5

  D      = \(\dfrac{5^{101}-5}{4}\)

Bài 2:

So sánh 

a, 54⁴ = (2.3³)⁴ = 2⁴.3¹²  

    21¹² = (3.7)¹² = 3¹².7¹²

Vì 2⁴ < 7¹² nên 54⁴ < 21¹²

b, 3³⁹ và 11²¹

    3³⁹   =   (3¹³)³

   11²¹ = (11⁷)³

     3¹³ = (3²)⁶.3 = 9⁶.3 < 9⁷ < 11⁷ 

Vậy 3³⁹  < 11²¹

1) \(D=5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=1+5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=\dfrac{5^{100+1}-1}{5-1}\)

\(\Rightarrow D+1=\dfrac{5^{101}-1}{4}\)

\(\Rightarrow D=\dfrac{5^{101}-1}{4}-1=\dfrac{5^{101}-5}{4}=\dfrac{5\left(5^{100}-1\right)}{4}\)

2)

a) \(21^{12}=\left(21^3\right)^4=9261^4>54^4\Rightarrow54^4< 21^{12}\)

b) \(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 11^{20}< 11^{21}\)

\(\Rightarrow3^{39}< 11^{21}\)

c) \(201^{60}=\left(201^4\right)^{15}=\text{1632240801}^{15}\)

\(398^{45}=\left(398^3\right)^{15}=\text{63044792}^{15}< \text{1632240801}^{15}\)

\(201^{60}>398^{45}\)

\(A=5^3+5^4+5^5+...+5^{100}\)

\(A=5^3\left(1+5^1+5^2+...+5^{97}\right)\)

\(A=5^3.\dfrac{5^{97+1}-1}{5-1}=\dfrac{5^3}{4}.\left(5^{98}-1\right)\)

5

Bài 2: 

Ta có: (x-3)(x+4)>0

=>x>3 hoặc x<-4

Bài 3:

a: \(5S=5-5^2+...+5^{99}-5^{100}\)

\(\Leftrightarrow6S=1-5^{100}\)

hay \(S=\dfrac{1-5^{100}}{6}\)

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:

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

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

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

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

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

\(A=5^2+5^4+5^6+...+5^{100}+5^{102}\\ =5^2.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)⋮25\)

\(=\left(5^2.1+5^2.5^2+5^2.5^4+....+5^2.5^{98}+5^2.5^{102}\right)\\ =5^2.\left(1+5^2+5^4+....+5^{98}+5^{102}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{102}\right)⋮25\\ =>A⋮25\)