Google
Câu hỏi: Cho $f,g$ là hai hàm số liên tục trên $\left[ 1;3 \right]$ thoả mãn điều kiện $\int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=10}$ đồng thời $\int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx=6.}$ Tính $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]dx.}$
A. $6.$
B. $7.$
C. $10.$
D. $8.$
A. $6.$
B. $7.$
C. $10.$
D. $8.$
Theo ta có $\left\{ \begin{aligned}
& \int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=10} \\
& \int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx=6} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)dx}+3\int\limits_{1}^{3}{g\left( x \right)dx}=10 \\
& 2\int\limits_{1}^{3}{f\left( x \right)dx}-\int\limits_{1}^{3}{g\left( x \right)dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)dx}=4 \\
& \int\limits_{1}^{3}{g\left( x \right)dx}=2 \\
\end{aligned} \right.$
suy ra $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]dx=\int\limits_{1}^{3}{f\left( x \right)dx}+}\int\limits_{1}^{3}{g\left( x \right)dx}=4+2=6.$
& \int\limits_{1}^{3}{\left[ f\left( x \right)+3g\left( x \right) \right]dx=10} \\
& \int\limits_{1}^{3}{\left[ 2f\left( x \right)-g\left( x \right) \right]dx=6} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)dx}+3\int\limits_{1}^{3}{g\left( x \right)dx}=10 \\
& 2\int\limits_{1}^{3}{f\left( x \right)dx}-\int\limits_{1}^{3}{g\left( x \right)dx}=6 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \int\limits_{1}^{3}{f\left( x \right)dx}=4 \\
& \int\limits_{1}^{3}{g\left( x \right)dx}=2 \\
\end{aligned} \right.$
suy ra $\int\limits_{1}^{3}{\left[ f\left( x \right)+g\left( x \right) \right]dx=\int\limits_{1}^{3}{f\left( x \right)dx}+}\int\limits_{1}^{3}{g\left( x \right)dx}=4+2=6.$
Đáp án A.
