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Câu hỏi: Cho hàm số $f\left(x \right)>0$ và có đạo hàm liên tục trên $\mathbb{R}$, thỏa mãn $\left(x+1 \right){f}'\left(x \right)=\frac{\sqrt{f\left(x \right)}}{x+2}$ và $f\left(0 \right)={{\left(\frac{\ln 2}{2} \right)}^{2}}$. Giá trị $f\left(3 \right)$ bằng
A. $\frac{1}{2}{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
B. $4{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
C. $\frac{1}{4}{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
D. $2{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
A. $\frac{1}{2}{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
B. $4{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
C. $\frac{1}{4}{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
D. $2{{\left(4\ln 2-\ln 5 \right)}^{2}}$.
Với $x\in \left[ 0;3 \right]$ ta có:
$\left( x+1 \right){f}'\left( x \right)=\frac{\sqrt{f\left( x \right)}}{x+2}$ $\Leftrightarrow \frac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=\frac{1}{\left( x+1 \right)\left( x+2 \right)}$
$\Rightarrow \int\limits_{0}^{3}{\frac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}\text{d}x}=\int\limits_{0}^{3}{\left( \frac{1}{x+1}-\frac{1}{x+2} \right)\text{d}x}$ $\Rightarrow \left. 2\sqrt{f\left( x \right)} \right|_{0}^{3}=\left. \ln \left| \frac{x+1}{x+2} \right| \right|_{0}^{3}$
$\Rightarrow 2\left( \sqrt{f\left( 3 \right)}-\sqrt{f\left( 0 \right)} \right)=\ln \frac{4}{5}-\ln \frac{1}{2}$ $\Rightarrow \sqrt{f\left( 3 \right)}-\sqrt{{{\left( \frac{\ln 2}{2} \right)}^{2}}}=\frac{1}{2}\ln \frac{8}{5}$
$\Rightarrow \sqrt{f\left( 3 \right)}=\frac{1}{2}\left( \ln \frac{8}{5}+\ln 2 \right)=\frac{1}{2}\left( 4\ln 2-\ln 5 \right)$ $\Rightarrow f\left( 3 \right)=\frac{1}{4}{{\left( 4\ln 2-\ln 5 \right)}^{2}}$.
Vậy $f\left( 3 \right)=\frac{1}{4}{{\left( 4\ln 2-\ln 5 \right)}^{2}}$.
$\left( x+1 \right){f}'\left( x \right)=\frac{\sqrt{f\left( x \right)}}{x+2}$ $\Leftrightarrow \frac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=\frac{1}{\left( x+1 \right)\left( x+2 \right)}$
$\Rightarrow \int\limits_{0}^{3}{\frac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}\text{d}x}=\int\limits_{0}^{3}{\left( \frac{1}{x+1}-\frac{1}{x+2} \right)\text{d}x}$ $\Rightarrow \left. 2\sqrt{f\left( x \right)} \right|_{0}^{3}=\left. \ln \left| \frac{x+1}{x+2} \right| \right|_{0}^{3}$
$\Rightarrow 2\left( \sqrt{f\left( 3 \right)}-\sqrt{f\left( 0 \right)} \right)=\ln \frac{4}{5}-\ln \frac{1}{2}$ $\Rightarrow \sqrt{f\left( 3 \right)}-\sqrt{{{\left( \frac{\ln 2}{2} \right)}^{2}}}=\frac{1}{2}\ln \frac{8}{5}$
$\Rightarrow \sqrt{f\left( 3 \right)}=\frac{1}{2}\left( \ln \frac{8}{5}+\ln 2 \right)=\frac{1}{2}\left( 4\ln 2-\ln 5 \right)$ $\Rightarrow f\left( 3 \right)=\frac{1}{4}{{\left( 4\ln 2-\ln 5 \right)}^{2}}$.
Vậy $f\left( 3 \right)=\frac{1}{4}{{\left( 4\ln 2-\ln 5 \right)}^{2}}$.
Đáp án C.