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Gâu gâu 

a,

`3A=3+3^3+3^3+...+3^{53}`

`3A-A=(3+3^3+3^3+...+3^{53})-(1+3+3^3+3^3+...+3^{52})`

`2A=3^{53}-1`

`A=(3^{53}-1)/2`

b,

`A=1+3+3^3+3^3+...+3^{52}`

`A=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^{50}+3^{51}+3^{52})`

`A=(1+3+3^2)+3^3*(1+3+3^2)+....+3^{50}*(1+3+3^2)`

`A=(1+3+3^2)*(1+3^3+....+3^{50})`

`A=13*(1+3^3+....+3^{50})`

Do `13 \vdots 13 => A=13*(1+3^3+....+3^{50})\vdots 13 `

Vậy `A \vdots 13 `

\(A=3+3^2+3^3+3...+3^{31}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)+3^{31}\\ =\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+...+3^{27}\left(3+3^2+3^3\right)+3^{31}\\ =39+3^3.39+...+3^{27}.39+3^{31}\\ =39.\left(1+3^3+...+3^{27}\right)+3^{31}\\ Mà:39.\left(1+3^3+...+3^{27}\right)⋮13\left(Do:39⋮13\right)\\ Mà:3^{31}:13\left(dư:3\right)\\ Vậy:39.\left(1+3^3+...+3^{27}\right)+3^{31}:13\left(dư:3\right)\\ \Rightarrow A:13\left(dư:3\right)\)

A = 3 + 3² + 3³ + ... + 3³¹

= 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷ + ... + 3²⁹ + 3³⁰ + 3³¹

= 3 + (3² + 3³ + 3⁴) + (3⁵ + 3⁶ + 3⁷) + ... + (3²⁹ + 3³⁰ + 3³¹)

= 3 + 3².(1 + 3 + 3²) + 3⁵.(1 + 3 + 3²) + ... + 3²⁹.(1 + 3 + 3²)

= 3 + 3².13 + 3⁵.13 + ... + 3²⁹.13

= 3 + 13.(3² + 3⁵ + ... + 3³¹)

Do 13.(3² + 3⁵ + ... + 3³¹) ⋮ 13

⇒ 3 + 13.(3² + 3⁵ + ... + 3³¹) chia 13 dư 3

Vậy A chia 13 dư 3

\(B=3+3^2+3^3+...+3^{100}\)

\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=3+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=3+3^2.13+...+3^{98}.13\)

\(=3+13\left(3^2+...+3^{98}\right)\)

\(\Rightarrow B⋮̸13\)

\(\Rightarrow B:13\) dư 3.

Các bạn giải nhanh giúp mình nhé. Mình cần gấp. Thanks!

Bạn ko biết gõ số mũ à gõ thế này bố ai mà hiểu được

\(A=1+3+3^2+3^3+...+3^{2022}\)

\(=1+\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)

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

\(=1+13\left(3+3^4+...+3^{2020}\right)\)

=>A chia 13 dư 1

Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$

$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$

$=1+13(3+3^4+...+3^{2014})$ 

$\Rightarrow A-1\vdots 13(1)$

Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$

$=1+40(3+....+3^{2013})$

$\Rightarrow A-1\vdots 5(2)$

Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$

$\Rightarrow A$ chia $65$ dư $1$

a: (x-3)(y+1)=15

=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}

=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}

b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)

=>m chia 13 dư 4

\(m=1+3+3^2+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

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

\(=1+40\left(3+3^5+...+3^{97}\right)\)

=>m chia 40 dư 1