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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left[ \dfrac{1}{3};3 \right]$ thỏa mãn $f\left( x \right)+x.f\left( \dfrac{1}{x} \right)={{x}^{3}}-x$. Giá trị của tích phân $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}}\text{d}x$ bằng
A. $\dfrac{8}{9}.$
B. $\dfrac{3}{4}.$
C. $\dfrac{16}{9}.$
D. $\dfrac{2}{3}.$
A. $\dfrac{8}{9}.$
B. $\dfrac{3}{4}.$
C. $\dfrac{16}{9}.$
D. $\dfrac{2}{3}.$
$f\left( x \right)+xf\left( \dfrac{1}{x} \right)={{x}^{3}}-x$ $\Leftrightarrow \dfrac{f\left( x \right)}{{{x}^{2}}+x}+\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}=x-1$ với $x\in \left[ \dfrac{1}{3};3 \right]$.
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}\text{d}x}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)\text{d}x}$ $\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{{{x}^{2}}}\dfrac{f\left( \dfrac{1}{x} \right)}{\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}\text{d}x}=\dfrac{16}{9}$
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}-\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{\dfrac{1}{x}+{{\left( \dfrac{1}{x} \right)}^{2}}}\text{d}\left( \dfrac{1}{x} \right)}=\dfrac{16}{9}$ $\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{f\left( t \right)}{t+{{t}^{2}}}\text{d}t}=\dfrac{16}{9}$
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}=\dfrac{16}{9}\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}=\dfrac{8}{9}$.
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}\text{d}x}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)\text{d}x}$ $\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{1}{{{x}^{2}}}\dfrac{f\left( \dfrac{1}{x} \right)}{\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}\text{d}x}=\dfrac{16}{9}$
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}-\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{\dfrac{1}{x}+{{\left( \dfrac{1}{x} \right)}^{2}}}\text{d}\left( \dfrac{1}{x} \right)}=\dfrac{16}{9}$ $\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}-\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{f\left( t \right)}{t+{{t}^{2}}}\text{d}t}=\dfrac{16}{9}$
$\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}+\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}=\dfrac{16}{9}\Leftrightarrow \int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}\text{d}x}=\dfrac{8}{9}$.
Đáp án A.