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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;+\infty \right)$ thỏa mãn $f\left( x \right)>0,\forall x\ge 0$ và $\left( x+1 \right){f}'\left( x \right)=\dfrac{\sqrt{f\left( x \right)}}{x+2},\forall x\ge 0.$ Tính $\sqrt{f\left( 2 \right)}-\sqrt{f\left( 1 \right)}.$
A. $\ln \dfrac{9}{8}.$
B. $\dfrac{1}{2}\ln \dfrac{9}{8}.$
C. $\ln \dfrac{4}{3}.$
D. $\dfrac{1}{2}\ln \dfrac{4}{3}.$
A. $\ln \dfrac{9}{8}.$
B. $\dfrac{1}{2}\ln \dfrac{9}{8}.$
C. $\ln \dfrac{4}{3}.$
D. $\dfrac{1}{2}\ln \dfrac{4}{3}.$
Ta có $\left( x+1 \right){f}'\left( x \right)=\dfrac{\sqrt{f\left( x \right)}}{x+2}\Rightarrow \dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=\dfrac{1}{\left( x+1 \right)\left( x+2 \right)}\Rightarrow {{\left( \sqrt{f\left( x \right)} \right)}^{\prime }}=\dfrac{1}{2\left( x+1 \right)\left( x+2 \right)}.$
Suy ra
$\int\limits_{1}^{2}{{{\left( \sqrt{f\left( x \right)} \right)}^{\prime }}\text{d}x}=\int\limits_{1}^{2}{\dfrac{1}{2\left( x+1 \right)\left( x+2 \right)}}\text{d}x\Rightarrow \sqrt{f\left( 2 \right)}-\sqrt{f\left( 1 \right)}=\dfrac{1}{2}\ln \dfrac{9}{8}.$
Suy ra
$\int\limits_{1}^{2}{{{\left( \sqrt{f\left( x \right)} \right)}^{\prime }}\text{d}x}=\int\limits_{1}^{2}{\dfrac{1}{2\left( x+1 \right)\left( x+2 \right)}}\text{d}x\Rightarrow \sqrt{f\left( 2 \right)}-\sqrt{f\left( 1 \right)}=\dfrac{1}{2}\ln \dfrac{9}{8}.$
Đáp án B.