`~rai~`
\(C_n^{n-2}+6n+5=A_{n+1}^2\quad(n\ge 2)\\\Leftrightarrow \dfrac{n!}{(n-2)!.(n-n+2)!}+6n+5=\dfrac{(n+1)!}{(n-1)!}\\\Leftrightarrow \dfrac{n(n-1)(n-2)!}{(n-2)!.2!}+6n+5=\dfrac{n(n+1)(n-1)!}{(n-1)!}\\\Leftrightarrow \dfrac{n(n-1)}{2}+6n+5=n(n+1)\\\Leftrightarrow n(n-1)+12n+10=2n(n+1)\\\Leftrightarrow n^2-n+12n+10=2n^2+2n\\\Leftrightarrow n^2-n+12n+10-2n^2-2n=0\\\Leftrightarrow -n^2+9n+10=0\quad(a-b+c=0)\\\Leftrightarrow \left[\begin{array}{I}x=10\quad\text{(thỏa mãn)}\\x=-1.\text{(không thỏa mãn)}\end{array}\right.\\\text{Vậy n=10.}\)
