Phương pháp giải:

\({\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{a^k}{b^{n - k}}} \).

Lời giải chi tiết:

\(\begin{array}{l}A_n^2 - C_n^3 = 10\,\,\left( {n \ge 3} \right) \Leftrightarrow \frac{{n!}}{{\left( {n - 2} \right)!}} - \frac{{n!}}{{3!\left( {n - 3} \right)!}} = 10 \Leftrightarrow n\left( {n - 1} \right) - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = 10\\ \Leftrightarrow 6{n^2} - 6n - {n^3} + 3{n^2} - 2n - 60 = 0 \Leftrightarrow  - {n^3} + 9{n^2} - 8n - 60 = 0 \Leftrightarrow \left[ \begin{array}{l}n =  - 2\,\,\left( {ktm} \right)\\n = 6\,\,\,\left( {tm} \right)\\n = 5\,\,\,\left( {tm} \right)\end{array} \right.\end{array}\)

Ta có \({\left( {{x^2} - \frac{2}{{{x^3}}}} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{{\left( {{x^2}} \right)}^{n - k}}.{{\left( { - \frac{2}{{{x^3}}}} \right)}^k}}  = \sum\limits_{k = 0}^n {C_n^k{{\left( { - 2} \right)}^k}{x^{2n - 5k}}} \)

Với \(n = 5 \Rightarrow {\left( {{x^2} - \frac{2}{{{x^3}}}} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{{\left( { - 2} \right)}^k}{x^{10 - 5k}}} \).

\(10 - 5k = 5 \Leftrightarrow k = 1 \Rightarrow {a_5} = C_5^1.{\left( { - 2} \right)^1} =  - 10\)

Với \(n = 6 \Rightarrow {\left( {{x^2} - \frac{2}{{{x^3}}}} \right)^6} = \sum\limits_{k = 0}^6 {C_6^k{{\left( { - 2} \right)}^k}{x^{12 - 5k}}} \)

\(12 - 5k = 5 \Leftrightarrow k = \frac{7}{5}\,\,\left( {ktm} \right)\).

Chọn D.