\(C=1+3+3^2+...+3^{11}\)

a) \(C=1+3+3^2+...+3^{11}\)

       \(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\)

       \(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+3^9\left(1+3+3^2\right)\)

      \(=13+3^3.13+3^6.13+3^9.13\)

      \(=13\left(1+3^3+3^6+3^9\right)⋮13\)

\(\Rightarrow C⋮13\)

b) \(C=1+3+3^2+...+3^{11}\)

       \(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

       \(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

      \(=40+3^4.40+3^8.40\)

      \(=40\left(1+3^4+3^8\right)⋮40\)

\(\Rightarrow C⋮40\)

C chia het cho ca 13 va 40
 

XY=15 X=10 Y=5

Là x nhân y chứ không phải xy gạch đầu bạn ạ

Mà bạn có thể giải chi tiết ra được không??????????????

\(x+2xy+3y=6\)

\(4xy+2x+6y=12\)

\(2x\left(2y+1\right)+3\left(2y+1\right)=12+3\)

\(\left(2x+3\right)\left(2y+1\right)=15\)

\(2x+3\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)

Mà 2x + 3 > 1 

Từ đó tìm được \(\left(x;y\right)\in\left\{\left(0;2\right),\left(1;1\right),\left(6;0\right)\right\}\)

TA CÓ:

\(A=4k+3=25a+17\left(a;k\in N\right)\Rightarrow A+33=4k+36=25a+50\)

\(\Rightarrow A+33⋮4;25\Rightarrow A+33⋮4.25\Rightarrow A+33⋮100\)

\(\Rightarrow A:100\left(duw100-33\right)=67\)