181.65+5².181- 3².181= 181.(65+25-9)= 181. 81=14661

(11²+12²+13²+14²+15²).(2³.27- 6³)= A. (2³.3³- 6³) =A.(6³-6³)=0

a) Ta có: 

\(A=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^0\cdot2^4}\)

\(A=\frac{3^{10}\left(11+5\right)}{1\cdot16}\)

\(A=\frac{3^{10}\cdot16}{16}=3^{10}\)

b) Ta có:

\(B=\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)

\(B=\frac{2^{10}\cdot13\cdot\left(1+5\right)}{2^2\cdot7\cdot2^3\cdot13}\)

\(B=\frac{2^{10}\cdot6\cdot13}{2^5\cdot7\cdot13}=\frac{2^{11}\cdot3\cdot13}{2^5\cdot7\cdot13}\)

\(B=\frac{2^6\cdot3}{7}=\frac{192}{7}\)

Bài j mà dễ v~ !

dễ thì bạn làm đi chớ

$\textbf{A)}$

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

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 $12$.**

Ta có:\(A=3+3^2+3^3+...+3^{17}\)

\(3A=3\cdot\left(3+3^2+3^3+...+3^{17}\right)\)

\(3A=3^2+3^3+3^4+...+3^{18}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{18}\right)-\left(3+3^2+3^3+...+3^{17}\right)\)

\(2A=3^{18}-3\)

\(A=\frac{3^{18}-3}{2}\)

Vì \(3^{18}-3>3^{18}-4\)

\(\Rightarrow\frac{3^{18}-3}{2}>\frac{3^{18}-4}{2}\)

\(\Rightarrow A>B\)

A = 3^{1 + 2 + 3 + 4 + 5 + 6 ... + 17}

A = 3¹⁵³

B = [ 3¹⁸ - 4 ]

Ta thấy rõ ràng A sẽ lớn hơn B vì 153 > 18 ( chưa kể phải trừ thêm 4 ở biểu thức B )

A > B

Phạm Nguyễn Tất Đạt đúng nhưng hơi dài dòng quá !!