Question

Ребят помогите пожалуйста спасите от двойки((подробно 

Answer 1

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Answer 2

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)

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