Câu hỏi 14

ĐK: \(n \le 5;\,\,n \in N\)

\(\dfrac{5}{{C_5^n}} - \dfrac{2}{{C_6^n}} = \dfrac{{14}}{{C_7^n}}\)

\( \Leftrightarrow \dfrac{{5.n!.\left( {5 - n} \right)!}}{{5!}} - \dfrac{{2.n!.\left( {6 - n} \right)!}}{{6!}} = \dfrac{{14.n!.\left( {7 - n} \right)!}}{{7!}}\)

\( \Leftrightarrow \dfrac{5}{{5!}} - \dfrac{{2\left( {6 - n} \right)}}{{6!}} = \dfrac{{14\left( {6 - n} \right)\left( {7 - n} \right)}}{{7!}}\)

\( \Leftrightarrow 5 - \dfrac{{6 - n}}{3} = \dfrac{{\left( {6 - n} \right)\left( {7 - n} \right)}}{3}\) (Nhân 2 vế với 5!)

\( \Leftrightarrow 15 - \left( {6 - n} \right) = \left( {6 - n} \right).\left( {7 - n} \right)\)

\( \Leftrightarrow {n^2} - 14n + 33 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 3\,\,\,\,\,\,\left( {tm} \right)\\n = 11\,\,\left( {ktm} \right)\end{array} \right.\)

Chọn D.