Ребят помогите пожалуйста спасите от двойки((подробно

A)
$\frac{(n+2)!(n^2-9)}{(n+4)!}=\frac{(n+2)!(n^2-9)}{(n+2)!(n+3)(n+4)}=\frac{(n-3)(n+3)}{(n+3)(n+4)} =\frac{n-3}{n+4}$

б)
$\frac{1}{(n-2)!}- \frac{n^3-n}{(n+1)!} =\frac{1}{(n-2)!}- \frac{n(n^2-1)}{(n-2)!(n-1)n(n+1)} =\frac{1}{(n-2)!}- \frac{n^2-1}{(n-2)!(n^2-1)} = \\\frac{1}{(n-2)!}- \frac{1}{(n-2)!}=0$

в)
$\frac{25m^5-m^3}{(5m+1)!}( \frac{1}{5(5m-2)!} )^{-1}= \frac{m^3(25m^2-1)}{(5m+1)!} 5(5m-2)!= \\ \frac{5(5m-2)!m^3(5m-1)(5m+1)}{(5m-2)!(5m-1)5m(5m+1)}= m^2$

г)
$\frac{(3k+3)!k!}{(3k)!}: \frac{(k+3)!(3k+1)}{3!(k^2+5k+6)}= \frac{(3k+3)!k!}{(3k)!} \frac{3!(k^2+5k+6)}{(k+3)!(3k+1)}$

k²+5k+6=0
D=5²-4*6=25-24=1
k₁=(-5-1)/2=-3 k₂=(-5+1)/2=-2
k²+5k+6=(k+3)(k+2)

$=\frac{(3k+3)!k!}{(3k)!} \frac{3!(k+3)(k+2)}{(k+3)!(3k+1)}=\frac{6(3k+3)!k!(k+3)(k+2)}{(3k)!(k+3)!(3k+1)}= \\ =\frac{6(3k+3)!k!(k+3)(k+2)}{(3k)!k!(k+1)(k+2)(k+3)(3k+1)}=\frac{6(3k+3)!}{(3k)!(k+1)(3k+1)}= \\=\frac{6(3k)!(3k+1)(3k+2)(3k+3)}{(3k)!(k+1)(3k+1)} =\frac{6(3k+2)(3k+3)}{(k+1)}=$
$=\frac{18(3k+2)(k+1)}{(k+1)}=18(3k+2)=54k+36$