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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left[ \dfrac{1}{2};2 \right]$ và thỏa mãn $f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x$. Tính tích phân $I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}$.
A. $I=\dfrac{1}{2}$
B. $I=\dfrac{3}{2}$
C. $I=\dfrac{5}{2}$
D. $I=\dfrac{7}{2}$
A. $I=\dfrac{1}{2}$
B. $I=\dfrac{3}{2}$
C. $I=\dfrac{5}{2}$
D. $I=\dfrac{7}{2}$
Từ giả thiết, thay x bằng $\dfrac{1}{x}$ ta được $f\left( \dfrac{1}{x} \right)+2f\left( x \right)=\dfrac{3}{x}$
Do đó ta có hệ $\left\{ \begin{aligned}
& f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x \\
& f\left( \dfrac{1}{x} \right)+2f\left( x \right)=\dfrac{3}{x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x \\
& 4f\left( x \right)+2f\left( \dfrac{1}{x} \right)=\dfrac{6}{x} \\
\end{aligned} \right.\Rightarrow f\left( x \right)=\dfrac{2}{x}-x$
Khi đó $I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}=\int\limits_{\dfrac{1}{2}}^{2}{\left( \dfrac{2}{{{x}^{2}}}-1 \right)dx}=\left( -\dfrac{2}{x}-x \right)\left| \begin{aligned}
& ^{2} \\
& _{\dfrac{1}{2}} \\
\end{aligned} \right.=\dfrac{3}{2}$
Do đó ta có hệ $\left\{ \begin{aligned}
& f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x \\
& f\left( \dfrac{1}{x} \right)+2f\left( x \right)=\dfrac{3}{x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( x \right)+2f\left( \dfrac{1}{x} \right)=3x \\
& 4f\left( x \right)+2f\left( \dfrac{1}{x} \right)=\dfrac{6}{x} \\
\end{aligned} \right.\Rightarrow f\left( x \right)=\dfrac{2}{x}-x$
Khi đó $I=\int\limits_{\dfrac{1}{2}}^{2}{\dfrac{f\left( x \right)}{x}dx}=\int\limits_{\dfrac{1}{2}}^{2}{\left( \dfrac{2}{{{x}^{2}}}-1 \right)dx}=\left( -\dfrac{2}{x}-x \right)\left| \begin{aligned}
& ^{2} \\
& _{\dfrac{1}{2}} \\
\end{aligned} \right.=\dfrac{3}{2}$
Đáp án B.