Question

Resolva as equações. se poder deixar os calculos agradeco

Answer 1

Resposta e explicação passo a passo:

$a)\ \frac{(n+1)!}{(n-1)!}=12\\\\Onde\ \frac{(n+1)!}{(n-1)!}=n(n+1),\ portanto:\\\\n(n+1)=12\\\\n^2+n=12\\\\n^2+n-12=0\\\\(n+4)(n-3)=0\\\\n=3\ ou\ n=-4$

$b)\ \frac{(n+10)!}{(n+8)!}=30\\\\Onde\ \frac{(n+10)!}{(n+8)!}=(n+10)(n+10),\ portanto:\\\\(n+10)(n+9)=30\\\\n^2+10n+9n+90=30\\\\n^2+19n+90-30=0\\\\n^2+19n+60=0\\\\(n+4)(n+15)=0\\\\n=-4\ ou\ n=-15$

$c)\ n!=1\\\\Como\ 0!=1\ e\ 1!=1,\ entao\ n=0\ ou\ n=1$