Lời giải chi tiết:

\(A_x^{10} + A_x^9 = 9A_x^8\)  \(\left( {x \ge 10;\,\,x \in {N^*}} \right)\)

\( \Leftrightarrow \dfrac{{x!}}{{\left( {x - 10} \right)!}} + \dfrac{{x!}}{{\left( {x - 9} \right)!}} = 9\dfrac{{x!}}{{\left( {x - 8} \right)!}}\)

\( \Leftrightarrow x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 9} \right) + x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 8} \right) = 9x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 7} \right)\)

\( \Leftrightarrow \left( {x - 8} \right)\left( {x - 9} \right) + \left( {x - 8} \right) = 9\)

\( \Leftrightarrow {x^2} - 9x - 8x + 72 + x - 8 - 9 = 0\)

\( \Leftrightarrow {x^2} - 16x + 55 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 11\,\,\,\,\left( {tm} \right)\\x = 5\,\,\,\,\,\left( {ktm} \right)\end{array} \right.\) .

Chọn B.