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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( x \right)+{f}'\left( x \right)={{e}^{-x}},\forall x\in \mathbb{R}$ và $f\left( 0 \right)=2$. Tất cả các nguyên hàm của $f\left( x \right){{e}^{2x}}$ là
A. $\left( x-2 \right){{e}^{x}}+{{e}^{x}}+C$
B. $\left( x+2 \right){{e}^{2x}}+{{e}^{x}}+C$
C. $\left( x-1 \right){{e}^{x}}+C$
D. $\left( x+1 \right){{e}^{x}}+C$
A. $\left( x-2 \right){{e}^{x}}+{{e}^{x}}+C$
B. $\left( x+2 \right){{e}^{2x}}+{{e}^{x}}+C$
C. $\left( x-1 \right){{e}^{x}}+C$
D. $\left( x+1 \right){{e}^{x}}+C$
HD: Ta có $f\left( x \right)+{f}'\left( x \right)={{e}^{-x}}\Leftrightarrow {{\left( {{e}^{x}} \right)}^{\prime }}.f\left( x \right)+{{e}^{x}}.{f}'\left( x \right)=1\Leftrightarrow {{\left( {{e}^{x}}.f\left( x \right) \right)}^{\prime }}=1$
$\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$ mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right){{e}^{2x}}=\left( x+2 \right){{e}^{x}}\Rightarrow \int{\left( x+2 \right){{e}^{x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx=}\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x+1 \right){{e}^{x}}+C.$
$\Leftrightarrow {{e}^{x}}.f\left( x \right)=\int{dx}=x+C\Leftrightarrow f\left( x \right)=\dfrac{x+C}{{{e}^{x}}}$ mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\left( x+2 \right){{e}^{-x}}$
Do đó $f\left( x \right){{e}^{2x}}=\left( x+2 \right){{e}^{x}}\Rightarrow \int{\left( x+2 \right){{e}^{x}}dx}=\int{\left( x+2 \right){{e}^{x}}dx=}\left( x+2 \right){{e}^{x}}-\int{{{e}^{x}}dx}=\left( x+1 \right){{e}^{x}}+C.$
Đáp án D.