Lời giải chi tiết:

 \(3A_x^2 - A_{2x}^2 + 42 = 0\)\(\left( {x \ge 2;\,\,x \in N} \right)\)

\( \Leftrightarrow 3\dfrac{{x!}}{{\left( {x - 2} \right)!}} - \dfrac{{\left( {2x} \right)!}}{{\left( {2x - 2} \right)!}} + 42 = 0\)

\( \Leftrightarrow 3x\left( {x - 1} \right) - 2x\left( {2x - 1} \right) + 42 = 0\)

\( \Leftrightarrow 3{x^2} - 3x - 4{x^2} + 2x + 42 = 0\)

\( \Leftrightarrow  - {x^2} - x + 42 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 6\,\,\,\,\,\,\left( {tm} \right)\\x =  - 7\,\,\left( {ktm} \right)\end{array} \right.\)

Chọn B.