Giải phương trình: \(\frac{1}{2}A_{2x}^2 - A_x^2 = \frac{6}{x}C_x^3 + 10\).

ĐK: \(\left\{ {\begin{array}{*{20}{c}}{x \ge 3}\\{x \in \mathbb{N}}\end{array}} \right.\)

PT \( \Leftrightarrow \frac{1}{2}\frac{{2x!}}{{\left( {2x - 2} \right)!}} - \frac{{x!}}{{\left( {x - 2} \right)!}} = \frac{6}{x}\frac{{x!}}{{\left( {x - 3} \right)!3!}} + 10\)

\( \Leftrightarrow \frac{1}{2}\left( {2x - 1} \right)2x - \left( {x - 1} \right)x = \frac{1}{x}\left( {x - 2} \right)\left( {x - 1} \right)x + 10\)

\( \Leftrightarrow 2{x^2} - x - {x^2} + x = {x^2} - 3x + 2 + 10 \Leftrightarrow 3x = 12 \Leftrightarrow x = 4\)

Vậy x = 4.