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Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm trên $(0 ;+\infty)$ thỏa mãn $f(x)+2 x f^{\prime}(x)=3 x^2 \sqrt{x}$ và $f(1)=-2$.
Tính giá trị của $f\left(\dfrac{9}{4}\right)$.
A. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{96}$
B. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{192}$
C. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{64}$.
D. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{288}$.
Tính giá trị của $f\left(\dfrac{9}{4}\right)$.
A. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{96}$
B. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{192}$
C. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{64}$.
D. $f\left(\dfrac{9}{4}\right)=\dfrac{409}{288}$.
Với $x \in(0 ;+\infty)$, ta có: $f(x)+2 x f^{\prime}(x)=3 x^2 \sqrt{x}$
$
\begin{gathered}
\Leftrightarrow \dfrac{f(x)}{2 \sqrt{x}}+\sqrt{x} f^{\prime}(x)=\dfrac{3 x^2}{2} \Leftrightarrow(\sqrt{x} f(x))^{\prime}=\dfrac{3 x^2}{2} \\
\Rightarrow \sqrt{x} f(x)=\int \dfrac{3 x^2}{2} d x=\dfrac{x^3}{2}+C \\
\Rightarrow f(x)=\dfrac{x^2 \sqrt{x}}{2}+\dfrac{C}{\sqrt{x}}
\end{gathered}
$
$
\begin{aligned}
& \text { Mà } f(1)=-2 \Rightarrow \dfrac{1}{2}+C=-2 \Rightarrow C=\dfrac{-5}{2} \\
& \Rightarrow f(x)=\dfrac{x^2 \sqrt{x}}{2}-\dfrac{5}{2 \sqrt{x}} \Rightarrow f\left(\dfrac{9}{4}\right)=\dfrac{409}{192}
\end{aligned}
$
$
\begin{gathered}
\Leftrightarrow \dfrac{f(x)}{2 \sqrt{x}}+\sqrt{x} f^{\prime}(x)=\dfrac{3 x^2}{2} \Leftrightarrow(\sqrt{x} f(x))^{\prime}=\dfrac{3 x^2}{2} \\
\Rightarrow \sqrt{x} f(x)=\int \dfrac{3 x^2}{2} d x=\dfrac{x^3}{2}+C \\
\Rightarrow f(x)=\dfrac{x^2 \sqrt{x}}{2}+\dfrac{C}{\sqrt{x}}
\end{gathered}
$
$
\begin{aligned}
& \text { Mà } f(1)=-2 \Rightarrow \dfrac{1}{2}+C=-2 \Rightarrow C=\dfrac{-5}{2} \\
& \Rightarrow f(x)=\dfrac{x^2 \sqrt{x}}{2}-\dfrac{5}{2 \sqrt{x}} \Rightarrow f\left(\dfrac{9}{4}\right)=\dfrac{409}{192}
\end{aligned}
$
Đáp án B.