Question

Resolva a equação e determine o valor n, tal que

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{\dfrac{n!}{n + 1} - n! = \dfrac{4(n + 1)!}{9}}$

$\mathsf{\dfrac{n! - n!(n+1)}{n + 1} = \dfrac{4.n!(n + 1)}{9}}$

$\mathsf{\dfrac{n!(1 - (n+1))}{n + 1} = \dfrac{4.n!(n + 1)}{9}}$

$\mathsf{\dfrac{1 - (n+1)}{n + 1} = \dfrac{4(n + 1)}{9}}$

$\mathsf{\dfrac{-n}{n + 1} = \dfrac{4n + 4}{9}}$

$\mathsf{-9n = (4n + 4).(n + 1)}$

$\mathsf{-9n = 4n^2 + 4n + 4n + 4}$

$\mathsf{4n^2 + 17n + 4 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (17)^2 - 4.4.4}$

$\mathsf{\Delta = 289 - 64}$

$\mathsf{\Delta = 225}$

$\mathsf{n = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-17 \pm \sqrt{225}}{8} \rightarrow \begin{cases}\mathsf{n' = \dfrac{-17 + 15}{8} = -\dfrac{2}{8} = -\dfrac{1}{4}}\\\\\mathsf{n'' = \dfrac{-17 - 15}{8} = -\dfrac{32}{8} = -4}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{\emptyset\}}}}$