$\textbf{A)}$

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

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 $12$.

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

con khong biet

Sai hết :)

$\textbf{A)}$

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

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 $12$.**

b: \(B=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)

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

\(B=2+2^2+...+2^{60}\)

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

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