`a)A= 3 +3^2+ 3^3 + 3^4 + .... + 3^2020`

`A=(3+3^2)+( 3^3+3^4)...(3^2019+3^2020)`

`A=3.(1+3)+3^3. (1+3)+...+3^2019 .(1+3)`

`A=3.4+3^2. 4+...+3^2019. 4`

`A=4.(3+3^2+...+3^2019)`

Vì `4vdots4=>4.(3+3^2+...+3^2019)vdots4=>A vdots4`

`b)x+9 vdots x+2`

`=>(x+9)/(x+2)=(x+2+7)/(x+2)=(x+2)/(x+2)+7/(x+2)=1+7/(x+2)`

`=>7 vdots x+2`

`=>x+2 in Ư(7)={-1;1; -7; 7}`

Ta lập bảng:

\begin{array}{|c|c|c|}\hline \text{x+2}&\text{-1}&\text{1}&\text{-7}&\text{7}\\\hline \text{x}&\text{-3}&\text{-1}&\text{-9}&5\\\hline\end{array}

`=>x in {-3; -1; 5; -9}`

Vì `x in N=> x in {5}`

`A = 3 + 3^{2} + ...+3^{2020}`

`A = (3 + 3^{2}) + (3^{3} + 3^{4}) + ...+(3^{2019} + 3^{2020})`

`A = 3 . (1 + 3) + 3^{3} . (1 + 3) + ...+3^{2019} . (1 + 3)`

`A = 3 . 4 + 3^{3} . 4 + ...+3^{2019} . 4`

`A = 4 . (3 + 3^{3} +...+ 3^{2019})`

$\text{Vì}$ `4 \vdots 4` $\text{nên}$  `(3 + 3^{3} +...+ 3^{2019}) \vdots 4`

$\Rightarrow$ `A = (3 + 3^{2} + ...+3^{2020}) \vdots 4`

_________________________

$\text{Ta có:}$ `x + 9 \vdots x + 2`

$\text{Mà:}$ `x + 2 \vdots x + 2`

$\text{Lập hiệu:}$ `(x + 9) - (x + 2) \vdots (x + 2)`

                             `= x + 9 - x - 2 \vdots (x + 2)`

                             `= (x - x) + (9 - 2) \vdots (x + 2)`

                             `= 7 \vdots (x + 2)`

$\text{Vì}$ `x \in \mathbb{N} => x + 2 \in \mathbb{N} `

`=> x + 2 \in Ư(7) = {1;7}`

$\text{Lập bảng giá trị}$

`x + 2`              `1`                        `7`

`x`                    `-1`                      `5`

$\text{Vậy}$ `x = 5`