Lời giải:

$C=1+5+5^2+5^4+.....+5^{98}+5^{100}$

$25C=5^2C=5^2+5^3+5^4+5^6+....+5^{100}+5^{102}$

$25C-C=(5^3+5^{102})-(5+1)$

$24C=5^{102}-119$

$C=\frac{5^{102}-119}{24}$

- x - 8 - 9=17
=> - x=17+8+9
=> - x=34
=> x= -34

-x=17+9+8

-x=34

x=-34

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?

\(a,5^6:5^4+2^3x2^2-1^{2017}\) 

 \(=5^2+2^5-1\)

\(=25+32-1\)

\(=57-1\)

\(=56\)

a) 5⁶ : 5⁴ + 2³ x 2² - 1²⁰¹⁷

= 5² + 2³ x 2² - 1

= 5² + 2^{5 - 1}

= 25 + 32 - 1

= 57 - 1

= 56

a) x + 22 + (-14) + 52 = x + 60

b) (-90) - (p + 10) + 100

= -90 - p - 10 + 100

= (-90 -10 + 100) -p  =  -p