a) Ta có: \(n^2+4n+3=\left(n+1\right)\left(n+3\right)\)
Mà n lẻ \(\Leftrightarrow n=2k+1\)( \(k\in Z\) )
\(\Leftrightarrow\left(n+1\right)\left(n+3\right)=\left(2k+1+1\right)\left(2k+1+3\right)\)
\(=\left(2k+2\right)\left(2k+4\right)\)
\(=4\left(k+1\right)\left(k+2\right)\)
Vì \(\left(k+1\right)\left(k+2\right)\) là tích 2 số nguyên liên tiếp nên \(\left(k+1\right)\left(k+2\right)⋮2\)
\(\Rightarrow4\left(k+1\right)\left(k+2\right)⋮4\cdot2=8\)( đpcm )
b) \(n^3+3n^2-n-3\)
\(=n^2\left(n+3\right)-\left(n+3\right)\)
\(=\left(n+3\right)\left(n^2-1\right)\)
\(=\left(n+3\right)\left(n-1\right)\left(n+1\right)\)
Vì n lẻ nên \(n=2p+1\) ( \(q\in Z\) )
Khi đó : \(\left(n+3\right)\left(n-1\right)\left(n+1\right)=\left(2p+1+3\right)\left(2q+1-1\right)\left(2q+1+1\right)\)
\(=\left(2q+4\right)\cdot2q\cdot\left(2q+2\right)\)
\(=8q\left(q+1\right)\left(q+2\right)\)
Vì \(q\left(q+1\right)\left(q+2\right)\) là tích 3 số nguyên liên tiếp nên \(\left\{{}\begin{matrix}q\left(q+1\right)\left(q+2\right)⋮3\\q\left(q+1\right)\left(q+2\right)⋮2\end{matrix}\right.\)
\(\Rightarrow q\left(q+1\right)\left(q+2\right)⋮3\cdot2=6\)
\(\Rightarrow8q\left(q+1\right)\left(q+2\right)⋮8\cdot6=48\)( đpcm )
\(n^3+3n^2-n-3=n^2\left(n+3\right)-\left(n+3\right)=\left(n^2-1\right)\left(n+3\right)=\left(n-1\right)\left(n+1\right)\left(n+3\right)\) n le => n=2k+1 \(\Rightarrow\left(n-1\right)\left(n+1\right)\left(n+3\right)=2k\left(2k+2\right)\left(2k+4\right)=8k\left(k+1\right)\left(k+2\right)\) k và k+1 là 2 stn liên tiếp =>\(k\left(k+1\right)⋮2\Rightarrow8k\left(k+1\right)⋮16\)
k;k+1;k+2 là 3 stn liên tiếp => \(k\left(k+1\right)\left(k+2\right)⋮3\Rightarrow n^3+3n^2-n-3⋮3.16=48\left(\left(3,16\right)=48\right)\)
n le => n=2k+1
\(n^2+4n+3=\left(2k+1\right)\left(2k+5\right)+3=4k^2+12k+5+3=4k^2+12k+8=4k\left(k+3\right)+8.\) k và k+3 là 2 sô ko cungf chan le
\(\Rightarrow k\left(k+3\right)⋮2\Rightarrow4k\left(k+3\right)⋮8\Rightarrow4k\left(k+3\right)+8⋮8\)