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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 $f'\left( x \right)=\dfrac{1}{{{x}^{2}}+x-2};\ f\left( -3 \right)-f\left( 3 \right)=0$ và $f\left( 0 \right)=\dfrac{1}{3}$. Giá trị của biểu thức $f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)$ bằng:
A. $\dfrac{1}{3}+\dfrac{1}{3}\ln 2.$
B. $1+\ln 80.$
C. $1+\ln 2+\dfrac{1}{3}\ln \dfrac{4}{5}.$
D. $1+\dfrac{1}{3}\ln \dfrac{8}{5}.$
A. $\dfrac{1}{3}+\dfrac{1}{3}\ln 2.$
B. $1+\ln 80.$
C. $1+\ln 2+\dfrac{1}{3}\ln \dfrac{4}{5}.$
D. $1+\dfrac{1}{3}\ln \dfrac{8}{5}.$
$\begin{aligned}
& f'\left( x \right)=\dfrac{1}{{{x}^{2}}+x-2}=\dfrac{1}{\left( x-1 \right)\left( x+2 \right)}=\dfrac{1}{3}\left( \dfrac{1}{x-1}-\dfrac{1}{x+2} \right) \\
& \Rightarrow f\left( x \right)=\dfrac{1}{3}\int{\left( \dfrac{1}{x-1}-\dfrac{1}{x+2} \right)dx}=\dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+C=\left\{ \begin{aligned}
& \dfrac{1}{3}\ln \dfrac{x-1}{x+2}+{{C}_{1}}\ khi\ x\in \left( -\propto ;-2 \right) \\
& \dfrac{1}{3}\ln \dfrac{1-x}{x+2}+{{C}_{2}}\ khi\ x\in \left( -2;1 \right) \\
& \dfrac{1}{3}\ln \dfrac{x-1}{x+2}+{{C}_{3}}\ khi\ x\in \left( 1;+\infty \right) \\
\end{aligned} \right. \\
\end{aligned}$
Do đó $f\left( -3 \right)-f\left( 3 \right)=0\Rightarrow \dfrac{1}{3}\ln 4+{{C}_{1}}-\dfrac{1}{3}\ln \dfrac{2}{5}-{{C}_{3}}=0\Leftrightarrow {{C}_{1}}-{{C}_{3}}=-\dfrac{1}{3}\ln 10.$
Và $f\left( 0 \right)=\dfrac{1}{3}\Rightarrow \dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{2}}=\dfrac{1}{3}\Rightarrow {{C}_{2}}=\dfrac{1}{3}+\dfrac{1}{3}\ln 2.$
Khi đó $\begin{aligned}
& f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)=\left( \dfrac{1}{3}\ln \dfrac{5}{2}+{{C}_{1}} \right)+\left( \dfrac{1}{3}\ln 2+{{C}_{2}} \right)-\left( \dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{3}} \right) \\
& =\dfrac{1}{3}\ln 10+\left( {{C}_{1}}-{{C}_{3}} \right)+{{C}_{2}}=\dfrac{1}{3}\ln 10-\dfrac{1}{3}\ln 10+\dfrac{1}{3}+\dfrac{1}{3}\ln 2=\dfrac{1}{3}+\dfrac{1}{3}\ln 2. \\
\end{aligned}$
& f'\left( x \right)=\dfrac{1}{{{x}^{2}}+x-2}=\dfrac{1}{\left( x-1 \right)\left( x+2 \right)}=\dfrac{1}{3}\left( \dfrac{1}{x-1}-\dfrac{1}{x+2} \right) \\
& \Rightarrow f\left( x \right)=\dfrac{1}{3}\int{\left( \dfrac{1}{x-1}-\dfrac{1}{x+2} \right)dx}=\dfrac{1}{3}\ln \left| \dfrac{x-1}{x+2} \right|+C=\left\{ \begin{aligned}
& \dfrac{1}{3}\ln \dfrac{x-1}{x+2}+{{C}_{1}}\ khi\ x\in \left( -\propto ;-2 \right) \\
& \dfrac{1}{3}\ln \dfrac{1-x}{x+2}+{{C}_{2}}\ khi\ x\in \left( -2;1 \right) \\
& \dfrac{1}{3}\ln \dfrac{x-1}{x+2}+{{C}_{3}}\ khi\ x\in \left( 1;+\infty \right) \\
\end{aligned} \right. \\
\end{aligned}$
Do đó $f\left( -3 \right)-f\left( 3 \right)=0\Rightarrow \dfrac{1}{3}\ln 4+{{C}_{1}}-\dfrac{1}{3}\ln \dfrac{2}{5}-{{C}_{3}}=0\Leftrightarrow {{C}_{1}}-{{C}_{3}}=-\dfrac{1}{3}\ln 10.$
Và $f\left( 0 \right)=\dfrac{1}{3}\Rightarrow \dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{2}}=\dfrac{1}{3}\Rightarrow {{C}_{2}}=\dfrac{1}{3}+\dfrac{1}{3}\ln 2.$
Khi đó $\begin{aligned}
& f\left( -4 \right)+f\left( -1 \right)-f\left( 4 \right)=\left( \dfrac{1}{3}\ln \dfrac{5}{2}+{{C}_{1}} \right)+\left( \dfrac{1}{3}\ln 2+{{C}_{2}} \right)-\left( \dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{3}} \right) \\
& =\dfrac{1}{3}\ln 10+\left( {{C}_{1}}-{{C}_{3}} \right)+{{C}_{2}}=\dfrac{1}{3}\ln 10-\dfrac{1}{3}\ln 10+\dfrac{1}{3}+\dfrac{1}{3}\ln 2=\dfrac{1}{3}+\dfrac{1}{3}\ln 2. \\
\end{aligned}$
Đáp án A.