Question

2) Simplifique as frações:

a) ꯭ ꯭n꯭!꯭ ꯭
(n - 1)!

b) ꯭ ꯭x!꯭ ꯭
(x - 2)!

c) ꯭ (n꯭ +꯭ 1꯭)!꯭
n!


d) ꯭ (2꯭x ꯭+ ꯭2)!꯭

(2x)!


e) ꯭x! ꯭(x ꯭+ ꯭2)!꯭
(x - 1)! (x + 1)!


f) ꯭ (n꯭ - ꯭1)! ꯭+ ꯭(n ꯭- 2꯭)!꯭

n!​

Answer 1

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$\boxed{\begin{array}{l}\rm a)~\sf\dfrac{n!}{(n-1)!}=\dfrac{n\cdot\diagup\!\!\!(n-\diagup\!\!\!1)!}{\diagup\!\!\!(n-\diagup\!\!\!1)!}=n\\\rm b)~\sf\dfrac{x!}{(x-2)!}=\dfrac{x\cdot(x-1)\cdot\diagup\!\!\!(x-\diagup\!\!\!2)!}{\diagup\!\!\!(x-\diagup\!\!\!2)!}=x\cdot(x-1)\\\rm c)~\sf\dfrac{(n+1)!}{n!}=\dfrac{(n+1)\cdot\diagup\!\!\!n!}{\diagup\!\!\!n!}=n+1\\\rm d)~\sf\dfrac{(2x+2)!}{(2x)!}=\dfrac{(2x+2)\cdot(2x+1)\cdot\diagdown\!\!\!\!\!\!2x!}{\diagdown\!\!\!\!\!\!2x!}=(2x+2)\cdot(2x+1)\end{array}}$

$\boxed{\begin{array}{l}\rm e)~\sf\dfrac{x!\cdot(x+2)!}{(x-1)!\cdot(x+1)!}=\dfrac{x\cdot\diagup\!\!\!(x-\diagup\!\!\!1)!\cdot(x+2)\cdot\diagup\!\!\!(x+\diagup\!\!\!\!1)!}{\diagup\!\!\!(x-\diagup\!\!\!1)!\cdot\diagup\!\!\!(x+\diagup\!\!\!\!1)!}\\\sf =x\cdot(x+2)\\\rm f)~\sf\dfrac{(n-1)!+(n-2)!}{n!}=\dfrac{(n-1)\cdot(n-2)!+(n-2)!}{n\cdot(n-1)\cdot(n-2)!}\\\sf=\dfrac{\diagup\!\!\!(n-\diagup!\!\!2)![n-\diagup\!\!\!1+\diagup\!\!\!1]}{n\cdot(n-1)\cdot\diagup\!\!\!(n-\diagup\!\!\!\!2)!}=\dfrac{n}{n}=1\end{array}}$