Question

Calcule N sabendo que:


c) (N+2)! • (N-1)!           6
-------------------------  =   ----
    (N+1)! • N!                5


d) (N-2)! = 2 • (N-4)!)
​

Answer 1

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{(n+2)!\cdot(n-1)!}{(n+1)!\cdot n!}=\dfrac{6}{5}\\\sf\dfrac{(n+2)\cdot\diagdown\!\!\!\!\!(n+\diagdown\!\!\!\!\!1)!\cdot\diagdown\!\!\!\!\!(n-\diagdown\!\!\!\!\!1)!}{\diagdown\!\!\!\!\!(n+\diagdown\!\!\!\!\!1)!\cdot n\cdot\diagdown\!\!\!\!\!(n-\diagdown\!\!\!\!\!1)!}=\dfrac{6}{5}\\\sf \dfrac{n+2}{n}=\dfrac{6}{5}\\\sf 6n=5n+10\\\sf 6n-5n=10\\\sf n=10\end{array}}$

$\boxed{\begin{array}{l}\sf (n-2)!=2\cdot(n-4)!\\\sf (n-2)\cdot(n-3)\cdot\diagdown\!\!\!\!(n-\diagdown\!\!\!\!\!4)!=2\cdot\diagdown\!\!\!\!\!(n-\diagdown\!\!\!\!\!4)!\\\sf n^2-5n+6-2=0\\\sf n^2-5n+4=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-5)^2-4\cdot1\cdot4\\\sf\Delta=25-16\\\sf\Delta=9\\\sf n=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf n=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot1}\\\sf n=\dfrac{ 5\pm3}{2}\begin{cases}\sf n_1=\dfrac{5+3}{2}=\dfrac{8}{2}=4\\\sf n_2=\dfrac{5-3}{2}=\dfrac{2}{2}=1\end{cases}\end{array}}$

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