Lời giải:
Đẳng thức tương đương:
$n!n(n-1)+72=6n(n-1)+12n!$
Đặt $n(n-1)=a; n!=b$ thì:
$ab+72=6a+12b$
$\Leftrightarrow (a-12)(6-b)=0$
$\Rightarrow a=12$ hoặc $b=6$
$\Leftrightarrow n(n-1)=12$ hoặc $n!=6$
Với $n(n-1)=12$ thì suy ra $n=4$
$n!=6\Rightarrow n=3$
Đáp án A.
\(P_nA_n^2+72=6\left(A_n^2+2P_n\right)\)
\(\Leftrightarrow n!.\dfrac{n!}{\left(n-2\right)!}+72=6\left(\dfrac{n!}{\left(n-2\right)!}+2n!\right)\)
\(\Leftrightarrow n!.\left[\dfrac{n!}{\left(n-2\right)!}-12\right]-6\left[\dfrac{n!}{\left(n-2\right)!}-12\right]=0\)
\(\Leftrightarrow\left[\dfrac{n!}{\left(n-2\right)!}-12\right]\left(n!-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{n!}{\left(n-2\right)!}=12\\n!=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(n-1\right)n=12\\n!=3!\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n^2-n-12=0\\n=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=4\\n=3\end{matrix}\right.\)
