Sử dụng phương pháp quy nạp 

Dùng sao hả bạn,giúp mk vói😢

Ta thấy : \(n\inℤ^+\Rightarrow n=k+1\left(k\inℕ\right)\)

Khi đó : \(A=2^{3\left(k+1\right)+1}+2^{3\left(k+1\right)-1}+1\)

\(=2^{3k+4}+2^{3k+2}+1\)

\(=8^k.16+8^k.4+1\equiv1.2+1.4+1\equiv0\left(mod7\right)\)

Do vậy : \(A⋮7\) mà \(A>7\forall n\inℤ^+\)

\(\Rightarrow\)\(A=2^{3n+1}+2^{3n-1}+1\) là hợp số (đpcm)

Do n nguyên dương, đặt \(n=m+1\) với m là số tự nhiên

\(\Rightarrow A=2^{3\left(m+1\right)-1}+2^{3\left(m+1\right)+1}+1=2^{3m+2}+2^{3\left(m+1\right)+1}+1\)

\(=4.8^m+2.8^{m+1}+1\)

Do \(8\equiv1\left(mod7\right)\Rightarrow\left\{{}\begin{matrix}8^m\equiv1\left(mod7\right)\\8^{m+1}\equiv1\left(mod7\right)\end{matrix}\right.\)

\(\Rightarrow4.8^m+2.8^{m+1}+1\equiv4+2+1\left(mod7\right)\)

\(\Rightarrow4.8^m+2.8^{m+1}+1⋮7\)

Đặt \(2n+1=a^2,3n+1=b^2\).

\(15n+8=9\left(2n+1\right)-\left(3n+1\right)=9a^2-b^2=\left(3a-b\right)\left(3a+b\right)\)

Hiển nhiên \(3a+b>1\).

Nếu \(3a-b=1\Rightarrow b+1⋮3\).

mà \(b^2\equiv1\left(mod3\right)\Leftrightarrow b\equiv1\left(mod3\right)\Leftrightarrow b\equiv2\left(mod3\right)\)mâu thuẫn

do đó \(3a-b\ne1\).

Do đó \(15n+8\)là hợp số. 

Ta có: n²+3n+5=n²+n+2n+5=n.(n+1)+2n+2+3=n.(n+1)+2.(n+1)+3=(n+2).(n+1)+2

Vì (n+2).(n+1) chia hết cho n+1.

=>(n+2).(n+1)+2 : n+1(dư 2)

Vậy n²+3n+5:n+1(dư 2)