cho mik hỏi câu này nữa   a= 2+2 mũ 3 + 2 mũ 5 +.....+2 mũ 51

$\textbf{A)}$

$A=\left(2^{\,2^{6n+2}}+3\right):7.$

Ta có: $2^3\equiv1\pmod7.$

Lại có $6n+2\ge2\Rightarrow2^{6n+2}$ chia hết cho $4$, nên $2^{6n+2}\equiv1\pmod3.$

Suy ra $2^{2^{6n+2}}\equiv2^1\equiv2\pmod7.$

Vậy $2^{2^{6n+2}}+3\equiv2+3\equiv5\pmod7.$

$\Rightarrow$ Số dư là $5$.

$\textbf{B)}$

$B=\left(2^{\,2^{3n+1}}+3\right):13.$

Ta có: $2^{12}\equiv1\pmod{13}.$

Lại có $2^{3n+1}=2\cdot8^n.$

Vì $8^2\equiv12,\;8^3\equiv5,\;8^4\equiv1\pmod{12}$ nên $8^n\equiv4\pmod{12}$ khi $n\equiv2\pmod4$, do đó $2^{3n+1}\equiv8\pmod{12}.$

Suy ra $2^{2^{3n+1}}\equiv2^8\equiv256\equiv9\pmod{13}.$

Vậy $2^{2^{3n+1}}+3\equiv9+3\equiv12\pmod{13}.$

$\Rightarrow$ **Số dư là $12$.**

a. \(625^5=\left(5^4\right)^5=5^{20}< 5^{21}=\left(5^3\right)^7=125^7\)

b. với n khác 0 \(3^{2n}=9^n>8^n=2^{3n}\)

Còn với n=0 thì \(3^{2n}=2^{3n}=1\)

a, \(A=2^{2^{6n+2}}\)

Ta có: \(2^{6n+2}\equiv1\left(mod3\right)\)

\(\Rightarrow2^{6n+2}=3k+1\left(k\in Z\right)\)

\(\Rightarrow A=2^{3k+1}=4.2^{3k}=4.8^k\equiv4.1\equiv4\left(mod7\right)\)

Vậy A chia 7 dư 4

$\textbf{B)}$

$B=\left(2^{\,2^{3n+1}}+3\right):13.$

Ta có: $2^{12}\equiv1\pmod{13}.$

Lại có $2^{3n+1}=2\cdot8^n.$

Vì $8^2\equiv12,\;8^3\equiv5,\;8^4\equiv1\pmod{12}$ nên $8^n\equiv4\pmod{12}$ khi $n\equiv2\pmod4$, do đó $2^{3n+1}\equiv8\pmod{12}.$

Suy ra $2^{2^{3n+1}}\equiv2^8\equiv256\equiv9\pmod{13}.$

Vậy $2^{2^{3n+1}}+3\equiv9+3\equiv12\pmod{13}.$

$\Rightarrow$ **Số dư là $12$.**

đơn giản

gamer

a, \(3^4\) 

b,  \(8^7:8^2\) 

c,  \(x^3.x^2.x\)  

d,  \(4^n.4^2\)  

e, \(3^{k+2}:3^k\)  

a, 3^4

b,8^7:8^2=8^5

c, x^3.x^2.x=x^6

d,4^n.4^2=4^(n+2)

e, 3^k+2:3^k

=3^k.(1+2)

=3^k.3

=3^(k+10

Mấy bài này đẽ ẹc mà !!!