Question

Resolva: (n!)^2 - 7.(n!) + 6 = 0.​

Answer 1

Resposta:

n=1 ou n=3

Explicação passo-a-passo:

(n!)²-7(n!)+6=0

Chamando x=n!

x²-7x+6=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-7x+6=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-7~e~c=6\\\\\Delta=(b)^{2}-4(a)(c)=(-7)^{2}-4(1)(6)=49-(24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-7)-\sqrt{25}}{2(1)}=\frac{7-5}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-7)+\sqrt{25}}{2(1)}=\frac{7+5}{2}=\frac{12}{2}=6\\\\S=\{1,~6\}$

Para x=1:

x=n!=1 => n=1, porque 1!=1

Para x=6:

x=n!=6 => n=3, porque 3!=3.2.1=6