Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l} + \,\,y' = \dfrac{{2.{e^{2x}}.x - {e^{2x}}}}{{{x^2}}} = \dfrac{{\left( {2x - 1} \right){e^{2x}}}}{{{x^2}}}\\ + \,\,y'' = \dfrac{{\left[ {\left( {2x - 1} \right).{e^{2x}}} \right]'{x^2} - {e^{2x}}\left( {2x - 1} \right)2x}}{{{x^4}}}\\\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{4{e^{2x}}.x.{x^2} - {e^{2x}}\left( {2x - 1} \right)2x}}{{{x^4}}}\\\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{4{e^{2x}}.{x^2} - 2.{e^{2x}}.\left( {2x - 1} \right)}}{{{x^3}}}\end{array}\)
\( + \) Xét đáp án A: \(2y' + xy'' = \dfrac{{2(2x - 1){e^{2x}}}}{{{x^2}}} + \dfrac{{x\left[ {4{e^{2x}}.{x^2} - 2{e^{2x}}.\left( {2x - 1} \right)} \right]}}{{{x^3}}}\)
\( = \dfrac{{2\left( {2x - 1} \right){e^{2x}} + 4{e^{2x}}.{x^2} - 2{e^{2x}}\left( {2x - 1} \right)}}{{{x^2}}} = \dfrac{{4{e^{2x}}.{x^2}}}{{{x^2}}} = 4{e^{2x}}\)
\( \Rightarrow \) Đáp án A đúng.
Chọn A
