جد قيمة n لكل مما يأتي:

$\frac{(n-1)!}{(n+1)!}=\frac{1}{6}$

$\begin{array}{rl}& \frac{(n-1)!}{(n+1)n(n-1)!}=\frac{1}{6}\Rightarrow \frac{1}{n(n+1)}=\frac{1}{6}\Rightarrow \frac{1}{n^2+n}=\frac{1}{6}\Rightarrow n^2+n=6\\ & n^2+n-6=0\Rightarrow (n+3)(n-2)=0\\ & \mathrm{إما} n+3=0\Rightarrow n=-3 \mathrm{تهمل}\Rightarrow \mathrm{أو} n-2=0\Rightarrow n=2\end{array}$

مضروب العدد

هو عملية ضرب تناني تنتهي بالواحد

$\begin{array}{rl}n!& =n(n-1)(n-2)(n-3)\dots \dots 1\\ 0!& =1\\ 1!& =1\\ 2!& =2\times 1=2\\ 3!& =3\times 2\times 1=6\\ 4!& =4\times 3\times 2\times 1=24\\ 5!& =5\times 4\times 3\times 2\times 1=120\\ 6!& =6\times 5\times 4\times 3\times 2\times 1=720\\ 7!& =7\times 6\times 5\times 4\times 3\times 2\times 1=5040\end{array}$

$\begin{array}{rl}& (n+2)!=(n+2)(n+1)!\\ & (n+1)!=(n+1)n!\\ & (n-1)!=(n-1)(n-2)!\end{array}$

1- جد قيمة n لكل مما يأتي:

- $\frac{n!}{(n-2)!}=6$

$\begin{array}{rl}& \frac{n(n-1)(n-2)!}{(n-2)!}=6\Rightarrow n(n-1)=6\\ & n^2-n-6=0\Rightarrow (n-3)(n+2)=0\\ & \mathrm{إما} n-3=0\Rightarrow n=3\\ & \mathrm{أو} n+2=0\Rightarrow n=-2 \mathrm{تهمل}\end{array}$

- $\frac{n!}{(n-2)!}=72$

$\begin{array}{rl}& \frac{n(n-1)(n-2)!}{(n-2)!}=72\Rightarrow n(n-1)=72\\ & n^2-n-72=0\Rightarrow (n-9)(n+8)=0\\ & \mathrm{إما} n-9=0\Rightarrow n=9\\ & \mathrm{أو} n+8=0\Rightarrow n=-8\end{array}$

- $\frac{(n-1)!}{(n+1)!}=\frac{1}{6}$

$\begin{array}{rl}& \frac{(n-1)!}{(n+1)n(n-1)!}=\frac{1}{6}\Rightarrow \frac{1}{n(n+1)}=\frac{1}{6}\Rightarrow \frac{1}{n^2+n}=\frac{1}{6}\Rightarrow n^2+n=6\\ & n^2+n-6=0\Rightarrow (n+3)(n-2)=0\\ & \mathrm{إما} n+3=0\Rightarrow n=-3 \mathrm{تهمل}\Rightarrow \mathrm{أو} n-2=0\Rightarrow n=2\end{array}$

- $\frac{(n+2)!}{n!}=6$

$\begin{array}{rl}& \frac{(n+2)(n+1)n!}{n!}=6\Rightarrow (n+2)(n+1)=6\\ & n^2+n+2n+2=6\Rightarrow n^2+3n+2-6=0\\ & n^2+3n-4=0\Rightarrow (n+4)(n-1)=0\\ & \mathrm{إما} n+4=0\Rightarrow n=-4 \mathrm{تهمل}\\ & \mathrm{أو} n-1=0\Rightarrow n=1\end{array}$

- $(n-1)!=6(n-3)!$

$\begin{array}{rl}& (n-1)(n-2)(n-3)!=6(n-3)!\\ & (n-1)(n-2)=6\\ & n^2-2n-n+2=6\Rightarrow n^2-3n+2-6=0\\ & n^2-3n-4=0\Rightarrow (n-4)(n+1)=0\\ & \mathrm{إما} n-4=0\Rightarrow n=4\Rightarrow \mathrm{أو} n+1=0\\ & n=-1 \mathrm{تهمل}\end{array}$

- $n!=6(n-2)!$

$\begin{array}{rl}& n(n-1)(n-2)!=6(n-2)!\Rightarrow n(n-1)=6\\ & n^2-n=6\Rightarrow n^2-n-6=0\\ & (n-3)(n+2)=0\Rightarrow \mathrm{إما} n-3=0\Rightarrow n=3\\ & \mathrm{أو} n+2=0\Rightarrow n=-2 \mathrm{تهمل}\end{array}$

- $n!=720$

$\begin{array}{rl}& n!=6\times 5\times 4\times 3\times 2\times 1\\ & n!=6!\\ & \therefore n=6\end{array}$

- $n!=120$

$\begin{array}{rl}& n!=5\times 4\times 3\times 2\times 1\\ & n!=5!\\ & \therefore n=5\end{array}$

ملاحظة:

إذا كان عدد = !n عند الحل تستخدم طريقة القسمة التصاعدية.