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Câu hỏi: Cho hàm số $y=\dfrac{{{e}^{2x}}}{2x}$, mệnh đề nào sau đây đúng?
A. $\dfrac{1}{2}y'-xy''={{e}^{2x}}.$
B. $\dfrac{1}{2}y'+xy''={{e}^{2x}}.$
C. $y'-\dfrac{1}{2}xy''={{e}^{2x}}.$
D. $y'+\dfrac{1}{2}xy''={{e}^{2x}}.$
A. $\dfrac{1}{2}y'-xy''={{e}^{2x}}.$
B. $\dfrac{1}{2}y'+xy''={{e}^{2x}}.$
C. $y'-\dfrac{1}{2}xy''={{e}^{2x}}.$
D. $y'+\dfrac{1}{2}xy''={{e}^{2x}}.$
Ta có $y'={{e}^{2x}}\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right);\ y''={{e}^{2x}}\left( \dfrac{2}{x}-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}} \right).$
Vậy $y'+\dfrac{1}{2}xy''={{e}^{2x}}\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right)+\dfrac{1}{2}x{{e}^{2x}}\left( \dfrac{2}{x}-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}} \right)={{e}^{2x}}.$
Vậy $y'+\dfrac{1}{2}xy''={{e}^{2x}}\left( \dfrac{1}{x}-\dfrac{1}{2{{x}^{2}}} \right)+\dfrac{1}{2}x{{e}^{2x}}\left( \dfrac{2}{x}-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}} \right)={{e}^{2x}}.$
Đáp án D.