Giá trị của $n \in \mathbb{N}$ bằng bao nhiêu, biết $\dfrac{5}{{C_5^n}} - \dfrac{2}{{C_6^n}} = \dfrac{{14}}{{C_7^n}}$.

* PP tự luận:

PT \( \Leftrightarrow \dfrac{5}{{\dfrac{{5!}}{{\left( {5 - n} \right)!n!}}}} - \dfrac{2}{{\dfrac{{6!}}{{\left( {6 - n} \right)!n!}}}} = \dfrac{{14}}{{\dfrac{{7!}}{{\left( {7 - n} \right)!n!}}}}\,\,,\,\,n \in \mathbb{N},0 \le n \le 5\)\( \Leftrightarrow \dfrac{{5.\left( {5 - n} \right)!n!}}{{5!}} - \dfrac{{2.\left( {6 - n} \right)!n!}}{{6!}} = \dfrac{{14.\left( {7 - n} \right)!n!}}{{7!}}\)\( \Leftrightarrow 5.6.7 - 2.7.\left( {6 - n} \right) = 14\left( {6 - n} \right)\left( {7 - n} \right)\)\( \Leftrightarrow 210 - 84 + 14n = 14{n^2} - 182n + 588\)\( \Leftrightarrow 14{n^2} - 196n + 462 = 0\)\( \Leftrightarrow \left[ \begin{array}{l}n = 11\,\left( L \right)\\n = 3\,\left( {TM} \right)\end{array} \right. \Leftrightarrow n = 3\).