Question

resolva as seguintes equações fatoriais:
A) $\left \((x+2)! \atop (x!) \right.$=0
B) $\left \(x!\atop (x-1)! \right.$=5!

Answer 1

a)

$\mathrm{\dfrac{(x+2)!}{x!}=0\ \to\ \dfrac{(x+2)(x+1)x!}{x!}=0}\\\\ \mathrm{(x+2)(x+1)=0\ \to\ x^2+x+2x+2=0}\\\\ \mathrm{x^2+3x+2=0\ \to\ a=1\ \| \ b=3\ \| \ c=2}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-3\pm\sqrt{3^2-4.1.2}}{2.1}}\\\\ \mathrm{x=\dfrac{-3\pm\sqrt{9-8}}{2}=\dfrac{-3\pm\sqrt{1}}{2}=\dfrac{-3\pm1}{2}}\\\\ \mathrm{x_1=\dfrac{-3+1}{2}=\dfrac{-2}{2}=-1\ \to\ \nexists\ x!}\\\\ \mathrm{x_2=\dfrac{-3-1}{2}=\dfrac{-4}{2}=-2\ \to\ \nexists\ x!}\\\\ \mathbf{x=\varnothing}$

b)

$\mathrm{\dfrac{x!}{(x-1)!}=5!\ \to\ \dfrac{x(x-1)!}{(x-1)!}=5!}\\\\ \mathrm{x=5!\ \to\ x=5.4.3.2.1}\ \to\ \mathbf{x=120}$