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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên tập xác định và thỏa mãn
$2f\left( x \right).\left[ 1-2{{x}^{2}}.f\left( x \right) \right]=x.{f}'\left( x \right);\text{ }f\left( 2 \right)=\dfrac{2}{3}$. Khi đó, $\int\limits_{1}^{3}{f\left( x \right).\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}$ bằng
A. 4.
B. 10.
C. $\dfrac{25}{2}$.
D. $\dfrac{21}{2}$.
$2f\left( x \right).\left[ 1-2{{x}^{2}}.f\left( x \right) \right]=x.{f}'\left( x \right);\text{ }f\left( 2 \right)=\dfrac{2}{3}$. Khi đó, $\int\limits_{1}^{3}{f\left( x \right).\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}$ bằng
A. 4.
B. 10.
C. $\dfrac{25}{2}$.
D. $\dfrac{21}{2}$.
Ta có $2f\left( x \right).\left[ 1-2{{x}^{2}}.f\left( x \right) \right]=x.{f}'\left( x \right)\Leftrightarrow 2f\left( x \right)-4{{x}^{2}}.{{f}^{2}}\left( x \right)=x.{f}'\left( x \right)$
$\Leftrightarrow 2x.f\left( x \right)-4{{x}^{3}}.{{f}^{2}}\left( x \right)={{x}^{2}}.{f}'\left( x \right)$
$\Leftrightarrow \dfrac{2x.f\left( x \right)-{{x}^{2}}.{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=4{{x}^{3}}$
$\Leftrightarrow {{\left( \dfrac{{{x}^{2}}}{f\left( x \right)} \right)}^{\prime }}=4{{x}^{3}}\Leftrightarrow \int{{{\left( \dfrac{{{x}^{2}}}{f\left( x \right)} \right)}^{\prime }}dx}=\int{4{{x}^{3}}dx}\Leftrightarrow \dfrac{{{x}^{2}}}{f\left( x \right)}={{x}^{4}}+C\Leftrightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{{{x}^{4}}+C}$.
Mà $f\left( 2 \right)=\dfrac{2}{3}\Leftrightarrow \dfrac{{{2}^{2}}}{{{2}^{4}}+C}=\dfrac{2}{3}\Leftrightarrow C=-10\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{{{x}^{4}}-10}$.
Ta có $\int\limits_{1}^{3}{f\left( x \right).\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}=\int\limits_{1}^{3}{\dfrac{{{x}^{2}}}{{{x}^{4}}-10}.\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}=4$.
$\Leftrightarrow 2x.f\left( x \right)-4{{x}^{3}}.{{f}^{2}}\left( x \right)={{x}^{2}}.{f}'\left( x \right)$
$\Leftrightarrow \dfrac{2x.f\left( x \right)-{{x}^{2}}.{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=4{{x}^{3}}$
$\Leftrightarrow {{\left( \dfrac{{{x}^{2}}}{f\left( x \right)} \right)}^{\prime }}=4{{x}^{3}}\Leftrightarrow \int{{{\left( \dfrac{{{x}^{2}}}{f\left( x \right)} \right)}^{\prime }}dx}=\int{4{{x}^{3}}dx}\Leftrightarrow \dfrac{{{x}^{2}}}{f\left( x \right)}={{x}^{4}}+C\Leftrightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{{{x}^{4}}+C}$.
Mà $f\left( 2 \right)=\dfrac{2}{3}\Leftrightarrow \dfrac{{{2}^{2}}}{{{2}^{4}}+C}=\dfrac{2}{3}\Leftrightarrow C=-10\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{{{x}^{4}}-10}$.
Ta có $\int\limits_{1}^{3}{f\left( x \right).\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}=\int\limits_{1}^{3}{\dfrac{{{x}^{2}}}{{{x}^{4}}-10}.\left( {{x}^{3}}-\dfrac{10}{x} \right)dx}=4$.
Đáp án A.