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Câu hỏi: Cho hàm số $y=f\left( x \right)$ thoả mãn $f\left( 0 \right)=-\dfrac{4}{3}$ và ${f}'\left( x \right)={{x}^{3}}{{f}^{2}}\left( x \right)$ với mọi $x\in \mathbb{R}$. Giá trị của $f\left( 3 \right)$ bằng
A. $-1.$
B. $-\dfrac{4}{19}.$
C. $-\dfrac{3}{4}.$
D. $-\dfrac{1}{21}.$
A. $-1.$
B. $-\dfrac{4}{19}.$
C. $-\dfrac{3}{4}.$
D. $-\dfrac{1}{21}.$
Ta có: ${f}'\left( x \right)={{x}^{3}}{{f}^{2}}\left( x \right)\Rightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}={{x}^{3}}$
Suy ra ${{\left[ -\dfrac{1}{f\left( x \right)} \right]}^{\prime }}={{x}^{3}}$. Do đó, $-\dfrac{1}{f\left( x \right)}=\int{{{x}^{3}}}\text{d}x=\dfrac{1}{4}{{x}^{4}}+C$
Do $f\left( 0 \right)=-\dfrac{4}{3}$ nên ta có $C=\dfrac{3}{4}$.
Suy ra $-\dfrac{1}{f\left( x \right)}=\dfrac{1}{4}{{x}^{4}}+\dfrac{3}{4}$. Do đó, $f\left( x \right)=\dfrac{-4}{{{x}^{4}}+3}$.
Khi đó, $f\left( 3 \right)=\dfrac{-4}{{{3}^{4}}+3}=-\dfrac{1}{21}$.
Suy ra ${{\left[ -\dfrac{1}{f\left( x \right)} \right]}^{\prime }}={{x}^{3}}$. Do đó, $-\dfrac{1}{f\left( x \right)}=\int{{{x}^{3}}}\text{d}x=\dfrac{1}{4}{{x}^{4}}+C$
Do $f\left( 0 \right)=-\dfrac{4}{3}$ nên ta có $C=\dfrac{3}{4}$.
Suy ra $-\dfrac{1}{f\left( x \right)}=\dfrac{1}{4}{{x}^{4}}+\dfrac{3}{4}$. Do đó, $f\left( x \right)=\dfrac{-4}{{{x}^{4}}+3}$.
Khi đó, $f\left( 3 \right)=\dfrac{-4}{{{3}^{4}}+3}=-\dfrac{1}{21}$.
Đáp án D.