Question

Trebuie rezolvate în N*​

Answer 1

$1)\,\,\,(n+2)!+(n+1)!\leq 50\cdot n!\\ \\\Rightarrow (n+1)!(n+2)+(n+1)!\leq 50\cdot n!\\ \\ \Rightarrow(n+1)!(n+2+1)\leq 50\cdot n!\\ \\\Rightarrow \dfrac{(n+1)!(n+3)}{n!}\leq 50\\ \\\Rightarrow (n+1)(n+3)\leq 50\\ \\ \Rightarrow n\in \{0,1,2,3,4,5\}$

$\\\\2)\,\,\,\dfrac{(n+1)!}{(n-1)!}\leq 6 \Rightarrow \dfrac{(n-1)!n(n+1)}{(n-1)!}\leq 6 \Rightarrow n(n+1)\leq 6\\ \\ \Rightarrow n\in \{1,2\}$

$\\\\3)\,\,\,A_{x}^2<40 \Rightarrow \dfrac{x!}{(x-2)!}<40 \Rightarrow \dfrac{(x-2)!(x-1)x}{(x-2)!}<40 \Rightarrow\\ \\ \Rightarrow (x-1)x<40 \Rightarrow x\in \{2,3,4,5,6\}$

$\\\\4)\,\,\,C_{x}^5\geq C_{x}^4\Rightarrow \dfrac{x!}{5!(x-5)!}\geq \dfrac{x!}{4!(x-4)!}\Big|:x! \Rightarrow\\ \\ \Rightarrow \dfrac{1}{4!\cdot 5\cdot (x-5)!}\geq \dfrac{1}{4!(x-5)!(x-4)}\Big|\cdot 4!(x-5)! \Rightarrow \\ \\ \Rightarrow \dfrac{1}{5}\geq \dfrac{1}{x-4} \Rightarrow 5\leq x-4 \Rightarrow x\geq 9\Rightarrow x\in \{9,10,11,12,...\}$