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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R},$ thỏa mãn điều kiện
$\left\{ \begin{aligned}
& 2x\left[ 1+f\left( x \right) \right]={{\left[ f'\left( x \right) \right]}^{3}},\forall x\in \mathbb{R} \\
& f\left( 0 \right)=-1 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{1}{4}.$
B. $-\dfrac{5}{6}.$
C. $\dfrac{1}{3}.$
D. $-\dfrac{2}{3}.$
$\left\{ \begin{aligned}
& 2x\left[ 1+f\left( x \right) \right]={{\left[ f'\left( x \right) \right]}^{3}},\forall x\in \mathbb{R} \\
& f\left( 0 \right)=-1 \\
\end{aligned} \right. $. Tính tích phân $ \int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{1}{4}.$
B. $-\dfrac{5}{6}.$
C. $\dfrac{1}{3}.$
D. $-\dfrac{2}{3}.$
HD: Ta có $2x\left[ 1+f\left( x \right) \right]={{\left[ f'\left( x \right) \right]}^{3}}\Leftrightarrow 2x=\dfrac{{{\left[ f'\left( x \right) \right]}^{3}}}{1+f\left( x \right)}\Leftrightarrow \sqrt[3]{2x}=\dfrac{f'\left( x \right)}{\sqrt[3]{1+f\left( x \right)}}$
$\Leftrightarrow \int{\dfrac{f'\left( x \right)}{\sqrt[3]{1+f\left( x \right)}}}dx=\int{\sqrt[3]{2x}}dx\Leftrightarrow \dfrac{3}{2}{{\left[ 1+f\left( x \right) \right]}^{\dfrac{3}{2}}}=\dfrac{3}{8}{{\left( 2x \right)}^{\dfrac{4}{3}}}+C$ mà $f\left( 0 \right)=-1\Rightarrow C=0.$
Vậy $f\left( x \right)=\dfrac{1}{8}.{{\left( 2x \right)}^{2}}-1\xrightarrow{{}}\int\limits_{0}^{1}{f\left( x \right)dx=-\dfrac{5}{6}.}$
$\Leftrightarrow \int{\dfrac{f'\left( x \right)}{\sqrt[3]{1+f\left( x \right)}}}dx=\int{\sqrt[3]{2x}}dx\Leftrightarrow \dfrac{3}{2}{{\left[ 1+f\left( x \right) \right]}^{\dfrac{3}{2}}}=\dfrac{3}{8}{{\left( 2x \right)}^{\dfrac{4}{3}}}+C$ mà $f\left( 0 \right)=-1\Rightarrow C=0.$
Vậy $f\left( x \right)=\dfrac{1}{8}.{{\left( 2x \right)}^{2}}-1\xrightarrow{{}}\int\limits_{0}^{1}{f\left( x \right)dx=-\dfrac{5}{6}.}$
Đáp án B.