Question

Resolva a equação:
(x+3)! + (x+2)! = 8 (x+1)!

Answer 1

Resposta: x = - 6; 0.

Explicação passo-a-passo:

É importante pensar na recursividade dos fatoriais.

$\mathtt{(n+3)!=(n+3)(n+2)(n+1)(n)(n-1)!}$

Com isso, podemos resolver:

$\mathtt{(x+3)!+(x+2)!=8(x+1)!}\\\\ \mathtt{(x+3)(x+2)(x+1)!+(x+2)(x+1)!=8(x+1)!}\\\\ \mathtt{\cancel{(x+1)!}\left[(x+3)(x+2)+(x+2)\right]=8\cancel{\left[(x+1)!\right]}}\\\\ \mathtt{(x+3)(x+2)+(x+2)=8}\\\\ \mathtt{(x^2+5x+6)+(x+2)=8}\\\\ \mathtt{(x^2+6x+8)=8}\\\\ \mathtt{x^2+6x+0=0}$

Finalizando:

$\mathtt{x^2+6x=0}\\\\ \mathtt{x(x+6)=0}\\\\ \therefore~\left\{\begin{array}{lcl} \mathtt{x+6=0}&\therefore&\mathtt{x=-6}\\\\ \mathtt{x=0} \end{array}\right.$