Đáp án: D
Giải thích các bước giải:
$\begin{array}{l}
y = \frac{{{e^{2x}}}}{{2x}}\\
\Rightarrow y' = \frac{1}{2}.\frac{{2.{e^{2x}}.x - {e^{2x}}}}{{{x^2}}} = \frac{{{e^{2x}}\left( {2x - 1} \right)}}{{2{x^2}}}\\
\Rightarrow y'' = \frac{1}{2}.\frac{{\left[ {2{e^{2x}}.\left( {2x - 1} \right) + 2.{e^{2x}}} \right].{x^2} - 2x.\left( {{e^{2x}}.2x - {e^{2x}}} \right)}}{{{x^4}}}\\
= \frac{{{e^{2x}}.\left( {2x - 1} \right).x + {e^{2x}}.x - {e^{2x}}.2x + {e^{2x}}}}{{{x^3}}}\\
= \frac{{{e^{2x}}}}{{{x^3}}}\\
A.\frac{1}{2}y' - x.y'' = \frac{1}{2}.\frac{{{e^{2x}}\left( {2x - 1} \right)}}{{2{x^2}}} - x.\frac{{{e^{2x}}}}{{{x^3}}}\\
= \frac{{{e^{2x}}.2x - {e^{2x}} - 4{e^{2x}}}}{{{x^2}}} \ne {e^{2x}}\\
B.\frac{1}{2}y' + x.y'' = \frac{1}{2}.\frac{{{e^{2x}}\left( {2x - 1} \right)}}{{2{x^2}}} + x.\frac{{{e^{2x}}}}{{{x^3}}}\\
= \frac{{{e^{2x}}.2x - {e^{2x}} + 4{e^{2x}}}}{{{x^2}}} \ne {e^{2x}}\\
C.y' - \frac{1}{2}xy'' = \frac{{{e^{2x}}\left( {2x - 1} \right)}}{{2{x^2}}} - \frac{1}{2}.x.\frac{{{e^{2x}}}}{{{x^3}}}\\
= \frac{{2x.{e^{2x}} - {e^{2x}} - {e^{2x}}}}{{2{x^2}}} = \frac{{2x{e^{2x}} + 2{e^{2x}}}}{{2{x^2}}}\\
D = y' + \frac{1}{2}xy'' = \frac{{{e^{2x}}\left( {2x - 1} \right)}}{{2{x^2}}} + \frac{1}{2}.x.\frac{{{e^{2x}}}}{{{x^3}}}\\
= \frac{{2x.{e^{2x}} - {e^{2x}} + {e^{2x}}}}{{2{x^2}}} = \frac{{2x{e^{2x}}}}{{2{x^2}}} = {e^{2x}}
\end{array}$
