a: \(G=8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

\(E=1+3+3^2+3^3+...+3^{1991}\)

\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)

\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)

A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259)  chia hết cho 3
=>A  chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7  chia hết cho 7 =>7.(2+...+258)  chia hết cho 7

CHIA HẾT CHO 3 :

A= (2+22)+(23+24)+...+(259+260)

A=2.(1+2)+23.(1+2)+...+259.(1+2)

A=2.3+23.3+...+259.3

A=3.(2+23+...+259)

Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3

=>A chia hết cho 3

dcv

llllllllllllllllllllllllllll

a) Ta có: \(32^{12}\cdot98^{20}\)

\(=2^{60}\cdot2^{20}\cdot7^{40}\)

\(=2^{80}\cdot7^{40}\)

\(=\left(2^2\cdot7\right)^{40}=28^{40}\)(đpcm)

b) Ta có: \(3^{1994}+3^{1993}-3^{1992}\)

\(=3^{1992}\left(3^2+3-1\right)\)

\(=3^{1992}\cdot11⋮11\)

x chia hết cho 2 và 250 < x < 260 là : 252 , 254 , 256 , 258

x chia hết cho 3 là và 480 < x < 490 là : 483 , 486 , 489

x vừa chia hết cho 2 vừa chia hết cho 5 và 300 < x < 350 là  : 310 , 320 , 330 , 340

x là số chẵn , x chịa hết cho 5 và 321 < x < 333 là : 330

                                 Chúc bạn học tốt .

\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^{259}+2^{260}\right)=6+6.2^2+6.2^4+...+6.2^{258}=3.2.\left(2^2+2^4+...+2^{258}\right)\)Chia hết cho 3

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

Ta có:

\(H=2+2^2+2^3+...+2^{60}\)

\(H=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(H=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(H=3\cdot\left(2+2^3+...+2^{59}\right)\)

Vậy H chia hết cho 3

_______

\(H=2+2^2+2^3+...+2^{60}\)

\(H=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(H=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(H=7\cdot\left(2+2^4+...+2^{58}\right)\)

Vậy H chia hết cho 7

__________

\(H=2+2^2+2^3+...+2^{60}\)

\(H=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(H=2\cdot\left(1+2+4+8\right)+2^5\cdot\left(1+2+4+8\right)+...+2^{57}\cdot\left(1+2+4+8\right)\)

\(H=15\cdot\left(2+2^5+...+2^{57}\right)\)

Vậy H chia hết cho 15 

$H=2+2^2+2^3+\mathrm{...}+2^{60}$

Ta có:

 $H=2.\left(1+2\right)+2^3.\left(1+2\right)+\mathrm{...}+2^{59}.(1+2)$

 $H=2.3+2^3.3+\mathrm{...}+2^{59}.3$

$H=3.\left(2+2^3+\mathrm{...}+2^{59}\right)⋮3$

Ta có:

 $H=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+\mathrm{...}+2^{28}.\left(1+2+2^2\right)$ 

Ta có:

 $H=2.\left(1+2+2^2+2^3\right)+2^5.\left(1+2+2^2+2^3\right)+\mathrm{...}+2^{57}.\left(1+2+2^2+2^3\right)$ 

$H=2.15+2^5.15+\mathrm{...}+2^{57}.15$

Vậy H chia hết cho  $3;7;15$.

nhớ tik đúng nha!!!