Tìm n thỏa mãn hệ thức sau:
a) $\frac{P_n-P_{n-1}}{P_{n+1}}=\frac16$                         b)$\frac{A^4_n}{A^3_{n+1}-C^{n-4}_n}=\frac{24}{23}$
a) Điều kiện: $n\geq 1,n\in Z^+$

VT có dạng:     $\dfrac{(n-1)!n-(n-1)!}{(n-1)!n(n+1)}=\dfrac{n-1}{n(n+1)}$

Ta có phương trình: $\dfrac{n-1}{n(n+1)}=\dfrac16\Leftrightarrow n^2-5n+6=0\Leftrightarrow \left[ \begin{array}{l}n_1=2\\n_2=3\end{array} \right.$
Đáp số: $n\in$ {$2;3$}

b) Điều kiện: $n\geq 4,n\in Z^+$

VT có dạng: $\dfrac{\frac{n!}{(n-4)!}}{\frac{(n+1)!}{(n-2)!}-\frac{n!}{(n-4)!4!}}=\dfrac{n!}{\frac{(n+1)!}{(n-3)(n-2)}-\frac{n!}{4!}}$

Đơn giản và rút gọn, ta có phương trình:
$24n^2-144n+120=0\Leftrightarrow n^2-6n+5=0\Leftrightarrow \left[ \begin{array}{l}n_1 = 1  (><)\\n_2 = 5\end{array} \right.$

Đáp số: $n=5$

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