Google
Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 2 \right)=-\dfrac{1}{5}$ và $f'\left( x \right)={{x}^{3}}{{\left[ f\left( x \right) \right]}^{2}}$ với mọi $x\in \mathbb{R}$. Giá trị của $f\left( 1 \right)$ bằng
A. $-\dfrac{4}{35}$
B. $-\dfrac{71}{20}$
C. $-\dfrac{79}{20}$
D. $-\dfrac{4}{5}$
A. $-\dfrac{4}{35}$
B. $-\dfrac{71}{20}$
C. $-\dfrac{79}{20}$
D. $-\dfrac{4}{5}$
Ta có: $f'\left( x \right)={{x}^{3}}{{\left[ f\left( x \right) \right]}^{2}}\Rightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}={{x}^{3}}\Rightarrow \int\limits_{1}^{2}{\dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}dx=\int\limits_{1}^{2}{{{x}^{3}}dx}}$
$\Leftrightarrow \left. \left( -\dfrac{1}{f\left( x \right)} \right) \right|_{1}^{2}=\dfrac{15}{4}\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+-\dfrac{1}{f\left( 1 \right)}=\dfrac{15}{4}\Leftrightarrow f\left( 1 \right)=-\dfrac{4}{5}$
$\Leftrightarrow \left. \left( -\dfrac{1}{f\left( x \right)} \right) \right|_{1}^{2}=\dfrac{15}{4}\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+-\dfrac{1}{f\left( 1 \right)}=\dfrac{15}{4}\Leftrightarrow f\left( 1 \right)=-\dfrac{4}{5}$
Đáp án D.