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Câu hỏi: Cho hàm số ${y=f\left( x \right)}$ có đạo hàm liên tục trên ${\left( 0 ; +\infty \right)}$ thỏa mãn ${\left( x+2 \right)f\left( x \right)=x{f}'\left( x \right)-{{x}^{3}}, \forall x\in \left( 0 ; +\infty \right)}$ và ${f\left( 1 \right)=e}$. Giá trị của ${f\left( 2 \right)}$ là
A. $4{{e}^{2}}+4e-2$.
B. $4{{e}^{2}}+4e-4$.
C. $4{{e}^{2}}+2e-2$.
D. $4{{e}^{2}}+2e-4$.
A. $4{{e}^{2}}+4e-2$.
B. $4{{e}^{2}}+4e-4$.
C. $4{{e}^{2}}+2e-2$.
D. $4{{e}^{2}}+2e-4$.
$\left( x+2 \right)f\left( x \right)=x{f}'\left( x \right)-{{x}^{3}}, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)-{{x}^{2}}f\left( x \right)={{x}^{4}}, \forall x\in \left( 0 ; +\infty \right)$ $\Leftrightarrow \dfrac{{{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)}{{{x}^{4}}}-\dfrac{f\left( x \right)}{{{x}^{2}}}=1, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}-\dfrac{f\left( x \right)}{{{x}^{2}}}=1, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}{{e}^{-x}}-\dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}}={{e}^{-x}}, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}} \right)}^{\prime }}={{e}^{-x}}, \forall x\in \left( 0 ; +\infty \right)$ $\Rightarrow \dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}}=-{{e}^{-x}}+C, \forall x\in \left( 0 ; +\infty \right)$.
Do $f\left( 1 \right)=e$, suy ra $C=1+{{e}^{-1}}$.
Vậy $f\left( x \right)={{x}^{2}}\left( -1+{{e}^{x}}+{{e}^{x-1}} \right), \forall x\in \left( 0 ; +\infty \right)$. Suy ra: $f\left( 2 \right)=4{{e}^{2}}+4e-4$.
$\Leftrightarrow {{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)-{{x}^{2}}f\left( x \right)={{x}^{4}}, \forall x\in \left( 0 ; +\infty \right)$ $\Leftrightarrow \dfrac{{{x}^{2}}{f}'\left( x \right)-2xf\left( x \right)}{{{x}^{4}}}-\dfrac{f\left( x \right)}{{{x}^{2}}}=1, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}-\dfrac{f\left( x \right)}{{{x}^{2}}}=1, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}} \right)}^{\prime }}{{e}^{-x}}-\dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}}={{e}^{-x}}, \forall x\in \left( 0 ; +\infty \right)$
$\Leftrightarrow {{\left( \dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}} \right)}^{\prime }}={{e}^{-x}}, \forall x\in \left( 0 ; +\infty \right)$ $\Rightarrow \dfrac{f\left( x \right)}{{{x}^{2}}}{{e}^{-x}}=-{{e}^{-x}}+C, \forall x\in \left( 0 ; +\infty \right)$.
Do $f\left( 1 \right)=e$, suy ra $C=1+{{e}^{-1}}$.
Vậy $f\left( x \right)={{x}^{2}}\left( -1+{{e}^{x}}+{{e}^{x-1}} \right), \forall x\in \left( 0 ; +\infty \right)$. Suy ra: $f\left( 2 \right)=4{{e}^{2}}+4e-4$.
Đáp án B.