Lời giải chi tiết:

Xét khai triển:

\({\left( {3x - 4} \right)^{17}} = C_{17}^0.{\left( {3x} \right)^{17}}.{\left( { - 4} \right)^0} + C_{17}^1.{\left( {3x} \right)^{16}}.{\left( { - 4} \right)^1} + C_{17}^2.{\left( {3x} \right)^{15}}.{\left( { - 4} \right)^2} + ... + C_{17}^{17}.{\left( {3x} \right)^0}.{\left( { - 4} \right)^{17}}\)

\( \Leftrightarrow {\left( {3x - 4} \right)^{17}} = C_{17}^0{.3^{17}}.{\left( { - 4} \right)^0}.{x^{17}} + C_{17}^1{.3^{16}}.{\left( { - 4} \right)^1}.{x^{16}} + ... + C_{17}^{17}{.3^0}.{\left( { - 4} \right)^{17}}.{x^0}\)

Thay \(x = 1\) vào, ta có:

\({\left( {3.1 - 4} \right)^{17}} = C_{17}^0{.3^{17}}.{\left( { - 4} \right)^0} + C_{17}^1{.3^{16}}.{\left( { - 4} \right)^1} + ... + C_{17}^{17}{.3^0}.{\left( { - 4} \right)^{17}}\)

\( \Leftrightarrow  - 1 = S\)\( \Leftrightarrow S =  - 1\)

Chọn B.