a, Ta có:

\(3^{2n+1}+2^{n+2}=9^n.3+2^n.4\)

\(=9^n.3-2^n.3+2^n.7=3\left(9^n-2^n\right)+2^n.7\)

Ta lại có:

\(9^n-2^n⋮9-2=7;2n.7⋮7\)

\(\Rightarrow3^{2n+1}+2^{n+2}⋮7\left(dpcm\right)\)

\(A=3^{n+2}-2^{n+3}+3^n-2^{n+2}\)

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

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

\(=3^n.10-2^{n+2}.3\)

Ta có:

\(3^n⋮3\) và \(10⋮2\) \(\Rightarrow\left(3^n.10\right)⋮6\)   (1)

\(2^{n+2}⋮2\) và \(3⋮3\Rightarrow\left(2^{n+2}.3\right)⋮6\)   (2)

Từ (1) và (2) \(\Rightarrow\left(3^n.10-2^{n+2}.3\right)⋮6\)

Vậy \(A⋮6\)

a) Giải:

Đặt \(A_n=11^{n+2}+12^{2n+1}\)\((*)\) Với \(n=0\) ta có:

\(A_0=11^2+12^1=133\) \(⋮133\Rightarrow\) \((*)\) đúng

Giả sử \((*)\) đúng đến giá trị \(k=n\) tức là:

\(B_k=11^{k+2}+12^{2k+1}\) \(⋮133\left(1\right)\)

Xét \(B_{k+1}-B_k\)

\(=11^{k+1+2}+12^{2\left(k+1\right)+1}-\left(11^{k+2}+12^{2k+1}\right)\)

\(=11^{k+3}-11^{k+2}+12^{2k+3}-12^{2k+1}\)

\(=10.11^{k+2}+143.12^{2k+1}\)

\(=10.121.11^k+143.12.144^k\)

\(\equiv\) \(10.121.11^k+10.12.11^k\)

\(\equiv\) \(10.11^k\left(121+12\right)\) \(\equiv\) \(0\left(mod133\right)\)

Theo giả thiết quy nạy \(\left(1\right)\) ta có: \(B_k⋮133\Leftrightarrow B_{k+1}⋮133\)

Hay \((*)\) đúng với \(n=k+1\) \(\Rightarrow\) Đpcm

_____________________Giải_____________________

\(\hept{\begin{cases}a+2b⋮3\\3a+3b⋮3\end{cases}}\Rightarrow3a+3b-a-2b⋮3\Rightarrow2a+b⋮3\)

2. _____________________Giải________________________

\(\hept{\begin{cases}a-b⋮7\\7a+7b⋮7\end{cases}}\Rightarrow7a+a+7b-b⋮7\Rightarrow8a+6b⋮7\)

=> 2(4a+3b) chia hết cho 7  vì  (2;7)=1

=> 4a+3b chia hết cho 7 (đpcm)