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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn các điều kiện $f\left( 1 \right)=2$, $f\left( x \right)\ne 0, \forall x>0$ và ${{\left( {{x}^{2}}+1 \right)}^{2}}f'\left( x \right)={{\left[ f\left( x \right) \right]}^{2}}\left( {{x}^{2}}-1 \right)$ với mọi $x>0$. Giá trị của $f\left( 2 \right)$ bằng
A. $\dfrac{2}{5}$.
B. $-\dfrac{2}{5}$.
C. $-\dfrac{5}{2}$.
D. $\dfrac{5}{2}$.
Lấy tích phân 2 vế trên $\left[ 1;2 \right]$ ta được
$\int\limits_{1}^{2}{\dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}\text{d}x}=\int\limits_{1}^{2}{\dfrac{{{x}^{2}}-1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}\text{d}x\Leftrightarrow -\dfrac{1}{f\left( x \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=\int\limits_{1}^{2}{\dfrac{1-\dfrac{1}{{{x}^{2}}}}{{{\left( x+\dfrac{1}{x} \right)}^{2}}}}\text{d}x$
$\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{f\left( 1 \right)}=\int\limits_{1}^{2}{\dfrac{\text{d}\left( x+\dfrac{1}{x} \right)}{{{\left( x+\dfrac{1}{x} \right)}^{2}}}}\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{2}=-\dfrac{1}{\left( x+\dfrac{1}{x} \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.$
$\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{2}=-\dfrac{2}{5}+\dfrac{1}{2}\Leftrightarrow f\left( 2 \right)=\dfrac{5}{2}$.
A. $\dfrac{2}{5}$.
B. $-\dfrac{2}{5}$.
C. $-\dfrac{5}{2}$.
D. $\dfrac{5}{2}$.
Ta có ${{\left( {{x}^{2}}+1 \right)}^{2}}f'\left( x \right)={{\left[ f\left( x \right) \right]}^{2}}\left( {{x}^{2}}-1 \right)\Leftrightarrow \dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}=\dfrac{{{x}^{2}}-1}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \forall x\in \left[ 1;2 \right] (*)$ Lấy tích phân 2 vế trên $\left[ 1;2 \right]$ ta được
$\int\limits_{1}^{2}{\dfrac{f'\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}\text{d}x}=\int\limits_{1}^{2}{\dfrac{{{x}^{2}}-1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}\text{d}x\Leftrightarrow -\dfrac{1}{f\left( x \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.=\int\limits_{1}^{2}{\dfrac{1-\dfrac{1}{{{x}^{2}}}}{{{\left( x+\dfrac{1}{x} \right)}^{2}}}}\text{d}x$
$\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{f\left( 1 \right)}=\int\limits_{1}^{2}{\dfrac{\text{d}\left( x+\dfrac{1}{x} \right)}{{{\left( x+\dfrac{1}{x} \right)}^{2}}}}\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{2}=-\dfrac{1}{\left( x+\dfrac{1}{x} \right)}\left| \begin{aligned}
& 2 \\
& 1 \\
\end{aligned} \right.$
$\Leftrightarrow -\dfrac{1}{f\left( 2 \right)}+\dfrac{1}{2}=-\dfrac{2}{5}+\dfrac{1}{2}\Leftrightarrow f\left( 2 \right)=\dfrac{5}{2}$.
Đáp án D.