Với $n$ thỏa mãn đẳng thức \(\dfrac{{A_n^4}}{{A_{n + 1}^3 - C_n^{n - 4}}} = \dfrac{{24}}{{23}}\) thì giá trị của biểu thức \(P = {\left( {n + 1} \right)^2} - 3n + 5\) là:

ĐK: \(n \in N;n \ge 4\)

$\begin{array}{l}\dfrac{{A_n^4}}{{A_{n + 1}^3 - C_n^{n - 4}}} = \dfrac{{24}}{{23}} \Leftrightarrow \dfrac{{A_n^4}}{{A_{n + 1}^3 - C_n^4}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{\left( {n + 1} \right)!}}{{\left( {n - 2} \right)!}} - \dfrac{{n!}}{{\left( {n - 4} \right)!4!}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{n!\left( {n + 1} \right)}}{{\left( {n - 4} \right)!\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{{n!}}{{\left( {n - 4} \right)!4!}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}\left( {\dfrac{{n + 1}}{{\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{1}{{24}}} \right)}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{1}{{\dfrac{{n + 1}}{{\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{1}{{24}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{24\left( {n - 3} \right)\left( {n - 2} \right)}}{{24n + 24 - \left( {n - 3} \right)\left( {n - 2} \right)}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow 23\left( {{n^2} - 5n + 6} \right) = 24n + 24 - \left( {{n^2} - 5n + 6} \right)\\ \Leftrightarrow 24{n^2} - 144n + 120 = 0 \Leftrightarrow \left[ \begin{array}{l}n = 5\,\,\left( {tm} \right)\\n = 1\,\,\,\,\left( {ktm} \right)\end{array} \right.\end{array}$

Với \(n = 5\) ta có: \(P = {\left( {5 + 1} \right)^2} - 3.5 + 5 = 26\)