Phương trình $c$, $d$ là một.
a,
$P_2.x^2-P_3.x=8$
$\to 2!.x^2-3!.x-8=0$
$\to 2x^2-6x-8=0$
$\to \left[ \begin{array}{l}x=-1\\x=4\end{array} \right.$
Vậy $S=\{-1; 4\}$
b,
ĐK: $x\ge 7; x\in\mathbb{N^*}$
$(x-7)!=720$
$\to (x-7)!=6!$
$\to x-7=6$
$\to x=13$ (TM)
Vậy $S=\{13\}$
c,
ĐK: $x\ge 1; x\in\mathbb{N^*}$
$\dfrac{P_x-P_{x-1}}{P_{x+1}}=\dfrac{1}{6}$
$\to \dfrac{x!-(x-1)!}{(x+1)!}=\dfrac{1}{6}$
$\to \dfrac{(x-1)!.x-(x-1)!}{(x-1)!.x(x+1)}=\dfrac{1}{6}$
$\to \dfrac{x-1}{x(x+1)}=\dfrac{1}{6}$
$\to x^2+x=6x-6$
$\to x^2-5x+6=0$
$\to \left[ \begin{array}{l}x=2\\x=3\end{array} \right.$ (TM)
Vậy $S=\{2;3\}$
d,
ĐK: $x\ge 1; x\in\mathbb{N^*}$
$\dfrac{x!-(x-1)!}{(x+1)!}=\dfrac{1}{6}$
$\to \dfrac{(x-1)!.x-(x-1)!}{(x-1)!.x(x+1)}=\dfrac{1}{6}$
$\to \dfrac{x-1}{x(x+1)}=\dfrac{1}{6}$
$\to x^2+x=6x-6$
$\to x^2-5x+6=0$
$\to \left[ \begin{array}{l}x=2\\x=3\end{array} \right.$ (TM)
Vậy $S=\{2;3\}$
