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Câu hỏi: Cho hai hàm số $f\left( x \right),g\left( x \right)$ liên tục trên đoạn $\left[ 1;2 \right]$ và thỏa mãn $\int\limits_{1}^{2}{\left[ 3f\left( x \right)+2g\left( x \right) \right]dx}=1,\int\limits_{1}^{2}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx}=-3.$ Khi đó, $\int\limits_{1}^{2}{f\left( x \right)dx}$ bằng:
A. $\dfrac{6}{7}$
B. $\dfrac{52}{3}$
C. $\dfrac{11}{7}$
D. $-\dfrac{5}{7}$
A. $\dfrac{6}{7}$
B. $\dfrac{52}{3}$
C. $\dfrac{11}{7}$
D. $-\dfrac{5}{7}$
Cách giải:
Ta có: $\left\{ \begin{aligned}
& \int\limits_{1}^{2}{\left[ 3f\left( x \right)+2g\left( x \right) \right]dx}=1 \\
& \int\limits_{1}^{2}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx}=-3 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 3\int\limits_{1}^{2}{f\left( x \right)dx}+2\int\limits_{1}^{2}{g\left( x \right)dx}=1 \\
& 2\int\limits_{1}^{2}{f\left( x \right)dx}-\int\limits_{1}^{2}{g\left( x \right)dx}=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{2}{f\left( x \right)dx}=-\dfrac{5}{7} \\
& \int\limits_{1}^{2}{g\left( x \right)dx}=\dfrac{11}{7} \\
\end{aligned} \right.$
Ta có: $\left\{ \begin{aligned}
& \int\limits_{1}^{2}{\left[ 3f\left( x \right)+2g\left( x \right) \right]dx}=1 \\
& \int\limits_{1}^{2}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx}=-3 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 3\int\limits_{1}^{2}{f\left( x \right)dx}+2\int\limits_{1}^{2}{g\left( x \right)dx}=1 \\
& 2\int\limits_{1}^{2}{f\left( x \right)dx}-\int\limits_{1}^{2}{g\left( x \right)dx}=-3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{2}{f\left( x \right)dx}=-\dfrac{5}{7} \\
& \int\limits_{1}^{2}{g\left( x \right)dx}=\dfrac{11}{7} \\
\end{aligned} \right.$
Đáp án D.