Lời giải:\(\begin{array}{l}{\left( {3 - 2x} \right)^{10}} = \sum\limits_{k = 0}^{10} {C_{10}^k{3^{n - k}}{{\left( { - 2} \right)}^k}.{x^k}} \\ \Rightarrow {a_0} = C_{10}^0{3^{10}}{\left( { - 2} \right)^0}\\\,\,\,\,\,\,\,{a_1} = C_{10}^1{3^9}{\left( { - 2} \right)^1}\\\,\,\,\,\,\,\,{a_2} = C_{10}^2{3^8}{\left( { - 2} \right)^2}\\\,\,\,\,\,\,\,....\\\,\,\,\,\,\,\,{a_{10}} = C_{10}^{10}{3^0}{\left( { - 2} \right)^{10}}\\ \Rightarrow S = {a_0} + {a_1} + ... + {a_{10}}\\\,\,\,\,\,\,\,\,\,\, = C_{10}^0{3^{10}}{\left( { - 2} \right)^0} + C_{10}^1{3^9}{\left( { - 2} \right)^1} + ... + C_{10}^{10}{3^0}{\left( { - 2} \right)^{10}}\\\,\,\,\,\,\,\,\,\,\, = {\left( {3 - 2} \right)^{10}} = {1^{10}} = 1.\end{array}\)
Chọn A.
