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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ -2;1 \right\}$ thỏa mãn $f\prime \left( x \right)=\dfrac{x-4}{{{x}^{2}}+x-2}$, $f\left( -3 \right)-f\left( 2 \right)=0$ và $f\left( 0 \right)=1$. Giá trị của biểu thức $f\left( -4 \right)+2f\left( -1 \right)-f\left( 3 \right)$ bằng
A. $3\ln \dfrac{5}{2}+2$.
B. $3\ln \dfrac{2}{5}+2$.
C. $2\ln \dfrac{2}{5}+2$.
D. $3\ln \dfrac{2}{5}+3$.
A. $3\ln \dfrac{5}{2}+2$.
B. $3\ln \dfrac{2}{5}+2$.
C. $2\ln \dfrac{2}{5}+2$.
D. $3\ln \dfrac{2}{5}+3$.
Ta có: $f\prime \left( x \right)=\dfrac{x-4}{{{x}^{2}}+x-2}=\dfrac{x-4}{\left( x+2 \right)\left( x-1 \right)}=\dfrac{2\left( x+1 \right)-1\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}=\dfrac{2}{x+2}+\dfrac{-1}{x-1}$
Suy ra: $f\left( x \right)=\int{{f}'\left( x \right)dx}=\left\{ \begin{matrix}
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{1}} & x<-2 \\
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{2}} & -2<x<1 \\
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{3}} & x>1 \\
\end{matrix} \right.$
$\Leftrightarrow f\left( x \right)=\left\{ \begin{matrix}
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{1-x}+{{C}_{1}}, & x<-2 \\
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{1-x}+{{C}_{2}}, & -2<x<1 \\
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{x-1}+{{C}_{3}}, & x>1 \\
\end{matrix} \right.$.
Lại có: $\left\{ \begin{aligned}
& f\left( -3 \right)-f\left( 2 \right)=0 \\
& f\left( 0 \right)=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \ln \dfrac{1}{4}+{{C}_{1}}-{{C}_{3}}-\ln 16=0 \\
& \ln 4+{{C}_{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{1}}-{{C}_{3}}=6\ln 2 \\
& {{C}_{2}}=1-2\ln 2 \\
\end{aligned} \right.$
Suy ra: $f\left( -4 \right)+2f\left( -1 \right)-f\left( 3 \right)=\ln \dfrac{4}{5}+{{C}_{1}}+2\left( \ln \dfrac{1}{2}+{{C}_{2}} \right)-\left( \ln \dfrac{25}{2}+{{C}_{3}} \right)$
$=\left[ \ln \dfrac{4}{5}+2\ln \dfrac{1}{2}-\ln \dfrac{25}{2} \right]+\left[ {{C}_{1}}+2{{C}_{2}}-{{C}_{3}} \right]=\ln \dfrac{2}{125}+6\ln 2+2\left( 1-2\ln 2 \right)=\ln \dfrac{8}{125}+2=3\ln \dfrac{2}{5}+2$.
Suy ra: $f\left( x \right)=\int{{f}'\left( x \right)dx}=\left\{ \begin{matrix}
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{1}} & x<-2 \\
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{2}} & -2<x<1 \\
2\ln \left| x+2 \right|-\ln \left| x-1 \right|+{{C}_{3}} & x>1 \\
\end{matrix} \right.$
$\Leftrightarrow f\left( x \right)=\left\{ \begin{matrix}
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{1-x}+{{C}_{1}}, & x<-2 \\
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{1-x}+{{C}_{2}}, & -2<x<1 \\
\ln \dfrac{{{\left( x+2 \right)}^{2}}}{x-1}+{{C}_{3}}, & x>1 \\
\end{matrix} \right.$.
Lại có: $\left\{ \begin{aligned}
& f\left( -3 \right)-f\left( 2 \right)=0 \\
& f\left( 0 \right)=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \ln \dfrac{1}{4}+{{C}_{1}}-{{C}_{3}}-\ln 16=0 \\
& \ln 4+{{C}_{2}}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{1}}-{{C}_{3}}=6\ln 2 \\
& {{C}_{2}}=1-2\ln 2 \\
\end{aligned} \right.$
Suy ra: $f\left( -4 \right)+2f\left( -1 \right)-f\left( 3 \right)=\ln \dfrac{4}{5}+{{C}_{1}}+2\left( \ln \dfrac{1}{2}+{{C}_{2}} \right)-\left( \ln \dfrac{25}{2}+{{C}_{3}} \right)$
$=\left[ \ln \dfrac{4}{5}+2\ln \dfrac{1}{2}-\ln \dfrac{25}{2} \right]+\left[ {{C}_{1}}+2{{C}_{2}}-{{C}_{3}} \right]=\ln \dfrac{2}{125}+6\ln 2+2\left( 1-2\ln 2 \right)=\ln \dfrac{8}{125}+2=3\ln \dfrac{2}{5}+2$.
Đáp án B.