Câu hỏi 33.2

\(y = \frac{{{{\left( {2x - 1} \right)}^9}}}{{{{\left( {x + 3} \right)}^8}}}\)

\(\begin{array}{l}y' = \frac{{9{{\left( {2x - 1} \right)}^8}.2.{{\left( {x + 3} \right)}^8} - {{\left( {2x - 1} \right)}^9}.8{{\left( {x + 3} \right)}^7}}}{{{{\left( {x + 3} \right)}^{16}}}}\\\,\,\,\,\,\, = \frac{{18{{\left( {2x - 1} \right)}^8}.\left( {x + 3} \right) - 8{{\left( {2x - 1} \right)}^9}}}{{{{\left( {x + 3} \right)}^9}}}\\\,\,\,\,\,\, = \frac{{2{{\left( {2x - 1} \right)}^8}.\left[ {9\left( {x + 3} \right) - 4\left( {2x - 1} \right)} \right]}}{{{{\left( {x + 3} \right)}^9}}}\\\,\,\,\,\,\, = \frac{{2{{\left( {2x - 1} \right)}^8}.\left( {x + 31} \right)}}{{{{\left( {x + 3} \right)}^9}}}\end{array}\)

Chọn C.