A = 3 + 3² + 3³ + 3⁴ + 3⁵+ .... + 3²⁰¹⁸ + 3²⁰¹⁹

   =  3 + (3² + 3³ + 3⁴ + 3⁵+ .... + 3²⁰¹⁸ + 3²⁰¹⁹)

   = 3 + [(3² + 3³) + (3⁴ + 3⁵) + ... + (3²⁰¹⁸ + 3²⁰¹⁹)]

   = 3 + [(3² + 3³) + 3².(3² + 3³) + ... + 3²⁰¹⁶.(3² + 3³)]

   = 3 + (36 + 3².36 + ... + 3²⁰¹⁶.36)

   = 3 + 36.(1 + 3² + .... + 3²⁰¹⁶)

   = 3 + 4.9.(1 + 3² + .... + 3²⁰¹⁶)

Vì  4.9.(1 + 3² + .... + 3²⁰¹⁶) \(⋮\)4

=> 4.9.(1 + 3² + .... + 3²⁰¹⁶) + 3 : 4 dư 3

=> A : 4 dư 3

Vậy số dư khi A chia 4 là 3

theo bài ra ta có:

  A=3^1+3^2+3^3+3^4 .... +3^2018+3^2019

3A=3.(3^1+3^2+3^3+3^4 .... +3^2018+3^2019)

3A=3^2+3^3+3^4 .... +3^2018+3^2020

3A-A=(3^2+3^3+3^4 .... +3^2018+3^2020)

        -(3^1+3^2+3^3+3^4 .... +3^2018+3^2019)

2A= 3^2020-3^1

=>2A=(...1)-(...3)

=>A=(...8)

...........

A=(1+2018)+2018^2(1+2018)+...+2018^2016(1+2018)

=2019(1+2018^2+...+2018^2016) chia hết cho 2019

=>A chia 2019 dư 0

AI NHANH MÌNH K , ĐANG CẦN GẤP

a)xét 2A =2+2^2+2^3+.....+2^2019

-A=1+2+2^2+...+2^2018

A=(2^2019)-1 <2^2019

b)theo câu a ta có A+1=2^2019-1+1=2^2019=2^(x+1)

2019=x+1 =>x=2018

c)theo câu b ta có A+1=2^2019=2.4^x=2^(1+2x)

=>2019=1+2x

tự làm nốt

\(A=1+7+7^2+7^3+...+7^{2019}+7^{2020}\\ \left(1+7+7^2\right)+7^3\left(1+7+7^2\right)+...+7^{2018}\left(1+7+7^2\right)\\ \left(1+7+7^2\right)\left(1+7^3+7^6+...+7^{2018}\right)\\ 57\left(1+7^3+7^6+...+7^{2018}\right)⋮57\)

A=1+7+7²+...+7²⁰¹⁹+7²⁰²⁰

=1+(7+7²+7³)+(7⁴+7⁵+7⁶)+...+(7²⁰¹⁸+7²⁰¹⁹+7²⁰²⁰)

=1+7(1+7+7²)+7⁴(1+7+7²)+...+7²⁰¹⁸(1+7+7²)

=1+7x57+7⁴x57+...+7²⁰¹⁸x57=1+57(7+7⁴+...+7²⁰¹⁸)

=>A chia cho 57 dư 1.vì 57(7+7⁴+...+7²⁰¹⁸)⋮57.

\(S=1+2-3-4+5+6-7-8+9-10-...+2018-2019-2020-2021\)

\(S=1+\left(2-3\right)-4+5+\left(6-7\right)-8+9-10-...+\left(2018-2019\right)-2020-2021\)

\(S=1-1+1-1+...-1-2020-2021=-1-2020-2021=-4042\)

b) Tích của số chia và thương là :

\(89-12=77\)=7.11

⇒ Số chia là 11; thương là 7

cộng 2021 nha bn

d

\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)

\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)

\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)

1) Ta có : 5xy + 2x - 5y = 7

=> x(5y - 2) - 5y + 2 = 7 + 2

=> x(5y - 2) - (5y - 2) = 9

=> (5y - 2)(x - 1) = 9

Với \(x;y\inℕ\Rightarrow\hept{\begin{cases}5y-2\inℕ^∗\\x-1\inℕ^∗\end{cases}}\)

=> có 9 = 3.3 = 1.9

Lập bảng xét các trường hợp 

Vậy x = 4 ; y = 1

2) A = 75.(4²⁰¹⁸ + 4²⁰¹⁷ + .... + 4² + 4) + 25

Đặt B = 4²⁰¹⁸ + 4²⁰¹⁷ + .... + 4² + 4 

Khi đó A = 75B + 25 

<=> 4B = 4²⁰¹⁹ + 4²⁰¹⁸ + .... + 4³ + 4²

Lấy 4B trừ B cả 2 vế ta có : 

4B - B = ( 4²⁰¹⁹ + 4²⁰¹⁸ + .... + 4³ + 4²) - (4²⁰¹⁸ + 4²⁰¹⁷ + .... + 4² + 4) 

   3B = 4²⁰¹⁹ - 4

=> B = \(\frac{4^{2019}-4}{3}\)

=> A = \(75\frac{4^{2019}-4}{3}+25=25.\left(4^{2019}-4\right)+25=25\left(4^{2019}-3\right)=25.4^{2019}-75\)

Vì \(25.4^{2019}⋮4^{2019}\Rightarrow25.4^{2019}-75:4^{2019}\text{ dư 75 }\Rightarrow A:4^{2019}\text{ dư 75}\)

Vậy số dư khi A chia cho 4²⁰¹⁹ là 75

\(A=1+2+2^2+...+2^{2019}+2^{2020}\)

\(A=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{2018}+2^{2019}+2^{2020}\right)\)

\(A=3+2^2\left(1+2+2^2\right)+...+2^{2018}\left(1+2+2^2\right)\)

\(A=3+2^2.7+....+2^{2018}.7\)

\(A=3+7\left(2^2+....+2^{2018}\right)\)

Vì 3 ko chia hết cho 7

=> A ko chia hết cho 7

=> A dư 3