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Câu hỏi: Biết $\int\limits_{0}^{3}{\dfrac{\text{d}x}{\left( x+2 \right)\left( x+4 \right)}}=a\ln 2+b\ln 5+c\ln 7$, $\left( a,b,c\in \mathbb{Q} \right)$. Giá trị của biểu thức $2a+3b-c$ bằng
A. $5$
B. $4$
C. $2$
D. $3$
A. $5$
B. $4$
C. $2$
D. $3$
$\int\limits_{0}^{3}{\dfrac{\text{d}x}{\left( x+2 \right)\left( x+4 \right)}}$ $=\dfrac{1}{2}\int\limits_{0}^{3}{\left( \dfrac{1}{x+2}-\dfrac{1}{x+4} \right)\text{d}x}$ $=\dfrac{1}{2}\left. \left( \ln \left| x+2 \right|-\ln \left| x+4 \right| \right) \right|_{0}^{3}$ $=\dfrac{1}{2}\ln 5-\dfrac{1}{2}\ln 7+\dfrac{1}{2}\ln 2$.
Khi đó: $2a+3b-c$ $=2.\dfrac{1}{2}+3.\dfrac{1}{2}+\dfrac{1}{2}=3$.
Khi đó: $2a+3b-c$ $=2.\dfrac{1}{2}+3.\dfrac{1}{2}+\dfrac{1}{2}=3$.
Đáp án D.