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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên $\mathbb{R}\backslash \left\{ 0 \right\}$ thỏa mãn ${f}'\left( x \right)+\dfrac{f\left( x \right)}{x}={{x}^{2}}$ và $f\left( 1 \right)=-1$. Giá trị của $f\left( \dfrac{3}{2} \right)$ bằng
A. $\dfrac{1}{96}$
B. $\dfrac{1}{64}$
C. $\dfrac{1}{48}$
D. $\dfrac{1}{24}$
A. $\dfrac{1}{96}$
B. $\dfrac{1}{64}$
C. $\dfrac{1}{48}$
D. $\dfrac{1}{24}$
Ta có
${f}'\left( x \right)+\dfrac{f\left( x \right)}{x}={{x}^{2}}\Leftrightarrow x.{f}'\left( x \right)+f\left( x \right)={{x}^{3}}\Leftrightarrow {{\left( x.f\left( x \right) \right)}^{\prime }}={{x}^{3}}\Rightarrow x.f\left( x \right)=\int{{{x}^{3}}}dx=\dfrac{{{x}^{4}}}{4}+C$
Vì $f\left( 1 \right)=-1\Rightarrow \dfrac{1}{4}+C=-1\Leftrightarrow C=-\dfrac{5}{4}\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{4}-\dfrac{5}{4x}\Rightarrow f\left( \dfrac{3}{2} \right)=\dfrac{1}{96}$
${f}'\left( x \right)+\dfrac{f\left( x \right)}{x}={{x}^{2}}\Leftrightarrow x.{f}'\left( x \right)+f\left( x \right)={{x}^{3}}\Leftrightarrow {{\left( x.f\left( x \right) \right)}^{\prime }}={{x}^{3}}\Rightarrow x.f\left( x \right)=\int{{{x}^{3}}}dx=\dfrac{{{x}^{4}}}{4}+C$
Vì $f\left( 1 \right)=-1\Rightarrow \dfrac{1}{4}+C=-1\Leftrightarrow C=-\dfrac{5}{4}\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{4}-\dfrac{5}{4x}\Rightarrow f\left( \dfrac{3}{2} \right)=\dfrac{1}{96}$
Đáp án A.