Question

8/(p+1)!+1/p! = 1/(p-1)!

Answer 1

$\frac{8}{(p+1)!}+ \frac{1}{p!}= \frac{1}{(p-1)!} \\ \\ \frac{8}{(p+1)p(p-1)!}+ \frac{1}{p(p-1)!}= \frac{1}{(p-1)!} ;mmc=(p-1)p(p-1)! \\ \\ 8+1(p+1)=p(p+1) \\ \\ 8+p+1=p^2+p \\ \\ p^2=9 \\ \\ p=-3;(\text{nao serve}) \\ ou \\ p=3$