Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=5+\left(5^2+5^3\right)+\left(5^4+5^5\right)+...+\left(5^{2020}+5^{2021}\right)\\ =5+5\left(5+5^2\right)+5^3\left(5+5^2\right)+...+5^{2019}\left(5+5^2\right)\\ =5+\left(5+5^2\right)\left(5+5^3+...+5^{2019}\right)\\ =5+31\left(5+5^3+...+5^{2019}\right)\)
Vậy BT chia 31 dư 5
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
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
Lời giải:
a. $(x-3)(y+1)=5=1.5=5.1=(-1)(-5)=(-5)(-1)$
Vì $x-3, y+1$ cũng là số nguyên nên ta có bảng sau:
b.
$A=21+5+(5^2+5^3)+(5^4+5^5)+....+(5^{98}+5^{99})$
$=26+5^2(1+5)+5^4(1+5)+....+5^{98}(1+5)$
$=2+24+(1+5)(5^2+5^4+...+5^{98}$
$=2+24+6(5^2+5^4+....+5^{98})=2+6(4+5^2+5^4+...+5^{98})$
$\Rightarrow A$ chia $6$ dư $2$.
\(A=5+5^2+5^3+...+5^{20}\)
\(A=\left(5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{18}+5^{19}+5^{20}\right)\)
\(A=30+5^3\cdot31+...+5^{18}\cdot31\)
\(A=30+31\cdot\left(5^3+5^6+...+5^{18}\right)\)
Mà: \(31\cdot\left(5^3+5^6+...+5^{18}\right)\) ⋮ 31
\(\Rightarrow A=30+31\cdot\left(5^3+5^6+...+5^{18}\right)\) chia cho 31 dư 30
A = 5 + 52 + 53 +...+ 520
A = 520 + 519 + 518 +...+ 53 + 52 + 5
A = (520 + 519 + 518) + (517 + 516 + 515) +...+ (55 + 54 + 53) + (52+ 5)
A = 518.( 52 + 5 + 1) + 515.(52 + 5 + 1) +...+ 53.(52+ 5 + 1) + (25 + 5)
A = 518. 31 + 515.31 +...+ 53.31 + 30
A = 31.(518 + 515 +...+ 53) + 30
31 ⋮ 31 ⇒ 31.(518 + 515 +...+53) ⋮ 31 mà 30 : 31 = 0 dư 31
Vậy A : 31 dư 30
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
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 \(S=\dfrac{1-5^{100}}{6}\)
A = 50 + 51 + 52 + 53 +...+5100 ( cs 101 so)
A = 50 +51 +( 52 + 53 + 54 )+( 55+56+57)+...+( 598 + 599 + 5100 )
A = 6+ 52.31 +55.31+...+598.31 chia 31 du 6
:)