Phương pháp giải:

Sử dụng các công thức:

\(\begin{array}{l}C_n^k = C_n^{n - k}\\C_n^k = \frac{{n!}}{{k!\left( {n - k} \right)!}}\,\,\left( {0 \le k \le n} \right)\end{array}\)

Lời giải chi tiết:

Đk: \(\left\{ \begin{array}{l}x - 1 \ge x - 4\\x + 1 \ge 3\\x - 1 \ge 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 4\\x \in N\end{array} \right.\)

\(\begin{array}{l}{x^2}C_{x - 1}^{x - 4} = A_4^2C_{x + 1}^3 - xC_{x - 1}^3\\ \Leftrightarrow {x^2}C_{x - 1}^3 + xC_{x - 1}^3 = A_4^2C_{x + 1}^3\\ \Leftrightarrow \left( {{x^2} + x} \right)C_{x - 1}^3 = 12C_{x + 1}^3\\ \Leftrightarrow \left( {{x^2} + x} \right)\frac{{\left( {x - 1} \right)!}}{{3!\left( {x - 4} \right)!}} = 12\frac{{\left( {x + 1} \right)!}}{{3!\left( {x - 2} \right)!}}\\ \Leftrightarrow \left( {{x^2} + x} \right) = 12\frac{{x\left( {x + 1} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\ \Leftrightarrow x\left( {x + 1} \right)\left( {x - 3} \right)\left( {x - 2} \right) - 12x\left( {x + 1} \right) = 0\\ \Leftrightarrow x\left( {x + 1} \right)\left( {{x^2} - 5x + 6 - 12} \right) = 0\\ \Leftrightarrow x\left( {x + 1} \right)\left( {{x^2} - 5x - 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x =  - 1\\x = 6\end{array} \right.\end{array}\)

Vậy x = 6.

Chọn C.