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Câu hỏi: Biết $\int\limits_{0}^{1}{{{\left( \dfrac{x-1}{x+2} \right)}^{2}}dx}=a+b\ln 2+c\ln 3\text{ }(a,b,c\in \mathbb{Q})$. Đẳng thức nào sau đây là đúng?
A. $2(a+b+c)=7$
B. $2(a+b-c)=7$
C. $2(a+b-c)=5$
D. $2(a+b+c)=5$
A. $2(a+b+c)=7$
B. $2(a+b-c)=7$
C. $2(a+b-c)=5$
D. $2(a+b+c)=5$
Ta có $\int\limits_{0}^{1}{{{\left( \dfrac{x-1}{x+2} \right)}^{2}}dx}=\int\limits_{0}^{1}{{{\left( 1-\dfrac{3}{x+2} \right)}^{2}}dx}=\int\limits_{0}^{1}{\left[ 1-\dfrac{6}{x+2}+\dfrac{9}{{{\left( x+2 \right)}^{2}}} \right]dx}=\left( x-6\ln \left| x+2 \right|-\dfrac{9}{x+2} \right)\left| \begin{matrix}
^{1} \\
_{0} \\
\end{matrix} \right.$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left( \dfrac{x-1}{x+2} \right)}^{2}}dx}=\dfrac{5}{2}+6\ln 2-6\ln 3\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{5}{2} \\
& b=6,c=-6 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2\left( a+b+c \right)=5 \\
& 2\left( a+b-c \right)=29 \\
\end{aligned} \right.$
^{1} \\
_{0} \\
\end{matrix} \right.$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left( \dfrac{x-1}{x+2} \right)}^{2}}dx}=\dfrac{5}{2}+6\ln 2-6\ln 3\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{5}{2} \\
& b=6,c=-6 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& 2\left( a+b+c \right)=5 \\
& 2\left( a+b-c \right)=29 \\
\end{aligned} \right.$
Đáp án D.