Question

(n + 1)!= 20.
(n-1)!

Answer 1

Resposta:

S = {4}

Explicação passo-a-passo:

Answer 2

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$\boxed{\begin{array}{l}\sf\dfrac{(n+1)!}{(n-1)!}=20\\\sf\dfrac{(n+1)\cdot n\cdot\diagup\!\!\!\!\!\!(n-\diagup\!\!\!\!\!\!1)!}{\diagup\!\!\!\!\!\!(n-\diagup\!\!\!\!\!\!1)!}=20\\\sf n^2+n=20\\\sf n^2+n-20=0\\\sf\Delta=b^2-4ac\\\sf\Delta=1^2-4\cdot1\cdot(-20)\\\sf\Delta=1+80\\\sf\Delta=81\\\sf n=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf n=\dfrac{-1\pm\sqrt{81}}{2\cdot1}\\\sf n=\dfrac{-1\pm9}{2}\begin{cases}\sf n_1=\dfrac{-1+9}{2}=\dfrac{8}{2}=4\\\sf n_2=\dfrac{-1-9}{2}=-\dfrac{10}{2}=-5\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\sf como~n\in\mathbb{N},a~soluc_{\!\!,}\tilde ao~n=-5~n\tilde ao~serve.\\\sf portanto\\\huge\boxed{\boxed{\boxed{\boxed{\sf n=4}}}}\end{array}}$