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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và thỏa mãn $f\left( 0 \right)=1$, ${{\left( {f}'\left( x \right) \right)}^{3}}={{e}^{x}}{{\left( f\left( x \right) \right)}^{2}}, \forall x\in \mathbb{R}$.
Tính $f\left( 3 \right)$
A. $f\left( 3 \right)={{e}^{2}}$.
B. $f\left( 3 \right)={{e}^{3}}$.
C. $f\left( 3 \right)=e$.
D. $f\left( 3 \right)=1$.
Tính $f\left( 3 \right)$
A. $f\left( 3 \right)={{e}^{2}}$.
B. $f\left( 3 \right)={{e}^{3}}$.
C. $f\left( 3 \right)=e$.
D. $f\left( 3 \right)=1$.
Ta có: ${{\left( {f}'\left( x \right) \right)}^{3}}={{e}^{x}}{{\left( f\left( x \right) \right)}^{2}}, \forall x\in \mathbb{R}\Leftrightarrow {f}'\left( x \right)=\sqrt[3]{{{e}^{x}}}.\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}\Leftrightarrow \dfrac{{f}'\left( x \right)}{\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}}=\sqrt[3]{{{e}^{x}}}$
$\Rightarrow \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}}dx}=\int\limits_{0}^{3}{\sqrt[3]{{{e}^{x}}}dx}\Leftrightarrow \int\limits_{0}^{3}{\dfrac{1}{\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}}df\left( x \right)}=\int\limits_{0}^{3}{{{e}^{\dfrac{x}{3}}}dx}\Leftrightarrow \left. 3\sqrt[3]{f\left( x \right)} \right|_{0}^{3}=\left. 3{{e}^{\dfrac{x}{3}}} \right|_{0}^{3}$
$\sqrt[3]{f\left( 3 \right)}-\sqrt[3]{f\left( 0 \right)}=e-1\Leftrightarrow \sqrt[3]{f\left( 3 \right)}-1=e-1\Leftrightarrow f\left( 3 \right)={{e}^{3}}$.
$\Rightarrow \int\limits_{0}^{3}{\dfrac{{f}'\left( x \right)}{\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}}dx}=\int\limits_{0}^{3}{\sqrt[3]{{{e}^{x}}}dx}\Leftrightarrow \int\limits_{0}^{3}{\dfrac{1}{\sqrt[3]{{{\left( f\left( x \right) \right)}^{2}}}}df\left( x \right)}=\int\limits_{0}^{3}{{{e}^{\dfrac{x}{3}}}dx}\Leftrightarrow \left. 3\sqrt[3]{f\left( x \right)} \right|_{0}^{3}=\left. 3{{e}^{\dfrac{x}{3}}} \right|_{0}^{3}$
$\sqrt[3]{f\left( 3 \right)}-\sqrt[3]{f\left( 0 \right)}=e-1\Leftrightarrow \sqrt[3]{f\left( 3 \right)}-1=e-1\Leftrightarrow f\left( 3 \right)={{e}^{3}}$.
Đáp án B.