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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left( -2;+\infty \right)$ thỏa mãn $f\left( x \right)+2\left( x+2 \right){f}'\left( x \right)=\dfrac{1}{\sqrt{x+2}}$ và $f\left( 2 \right)=\dfrac{1}{4}\ln 4$. Giá trị của $f\left( 7 \right)$ bằng
A. $f\left( 7 \right)=\dfrac{1}{2}\ln 3+3$.
B. $f\left( 7 \right)=\dfrac{1}{3}\ln 3+\dfrac{1}{2}$.
C. $f\left( 7 \right)=\dfrac{1}{3}\ln 3+1$.
D. $f\left( 7 \right)=\dfrac{1}{3}\ln 3$.
A. $f\left( 7 \right)=\dfrac{1}{2}\ln 3+3$.
B. $f\left( 7 \right)=\dfrac{1}{3}\ln 3+\dfrac{1}{2}$.
C. $f\left( 7 \right)=\dfrac{1}{3}\ln 3+1$.
D. $f\left( 7 \right)=\dfrac{1}{3}\ln 3$.
Nhân cả 2 vế của phương trình với $\dfrac{1}{2\sqrt{x+2}}$ ta được:
$\begin{aligned}
& \dfrac{1}{2\sqrt{x+2}}f\left( x \right)+\sqrt{x+2}.{f}'\left( x \right)=\dfrac{1}{2\left( x+2 \right)} \\
& \Leftrightarrow {{\left( \sqrt{x+2}.f\left( x \right) \right)}^{\prime }}=\dfrac{1}{2\left( x+2 \right)} \\
& \Leftrightarrow \sqrt{x+2}.f\left( x \right)=\int{\dfrac{1}{2\left( x+2 \right)}dx} \\
& \Leftrightarrow \sqrt{x+2}.f\left( x \right)=\dfrac{1}{2}.\ln \left( x+2 \right)+C \\
\end{aligned}$
Với $x=2$ ta được:
$2.f\left( 2 \right)=\dfrac{1}{2}.\ln 4+C\Leftrightarrow 2.\dfrac{1}{4}\ln 4=\dfrac{1}{2}\ln 4+C\Leftrightarrow C=0$
Ta có: $\sqrt{x+2}.f\left( x \right)=\dfrac{1}{2}\ln \left( x+2 \right)$
Thay $x=7$ ta được:
$3.f\left( 7 \right)=\dfrac{1}{2}\ln 9\Leftrightarrow f\left( 7 \right)=\dfrac{1}{3}\ln 3$.
$\begin{aligned}
& \dfrac{1}{2\sqrt{x+2}}f\left( x \right)+\sqrt{x+2}.{f}'\left( x \right)=\dfrac{1}{2\left( x+2 \right)} \\
& \Leftrightarrow {{\left( \sqrt{x+2}.f\left( x \right) \right)}^{\prime }}=\dfrac{1}{2\left( x+2 \right)} \\
& \Leftrightarrow \sqrt{x+2}.f\left( x \right)=\int{\dfrac{1}{2\left( x+2 \right)}dx} \\
& \Leftrightarrow \sqrt{x+2}.f\left( x \right)=\dfrac{1}{2}.\ln \left( x+2 \right)+C \\
\end{aligned}$
Với $x=2$ ta được:
$2.f\left( 2 \right)=\dfrac{1}{2}.\ln 4+C\Leftrightarrow 2.\dfrac{1}{4}\ln 4=\dfrac{1}{2}\ln 4+C\Leftrightarrow C=0$
Ta có: $\sqrt{x+2}.f\left( x \right)=\dfrac{1}{2}\ln \left( x+2 \right)$
Thay $x=7$ ta được:
$3.f\left( 7 \right)=\dfrac{1}{2}\ln 9\Leftrightarrow f\left( 7 \right)=\dfrac{1}{3}\ln 3$.
Đáp án D.