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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\!\!\backslash\!\!\left\{ -2 ; 2 \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{4}{{{x}^{2}}-4}$, $f\left( -3 \right)+f\left( 3 \right)=f\left( -1 \right)+f\left( 1 \right)=2$. Giá trị biểu thức $f\left( -4 \right)+f\left( 0 \right)+f\left( 4 \right)$ bằng
A. $4$.
B. $1$.
C. $2$.
D. $3$.
A. $4$.
B. $1$.
C. $2$.
D. $3$.
Ta có: $\int{\dfrac{4}{{{x}^{2}}-4} \text{d}x}=\int{\left( \dfrac{1}{x-2}-\dfrac{1}{x+2} \right) \text{d}x}=\ln \left| x-2 \right|-\ln \left| x+2 \right|+C$.
Do đó: $f\left( x \right)=\left\{ \begin{aligned}
& \ln \dfrac{x-2}{x+2}+{{C}_{1}} \text{khi} x<-2 \\
& \ln \dfrac{2-x}{x+2}+{{C}_{2}} \text{khi} -2<x<2 \\
& \ln \dfrac{x-2}{x+2}+{{C}_{3}} \text{khi} x>2 \\
\end{aligned} \right.$
$f\left( -3 \right)=\ln 5+{{C}_{1}};$ $f\left( 3 \right)=\ln \dfrac{1}{5}+{{C}_{3}};$ $f\left( 0 \right)={{C}_{2}}$ ; $f\left( -1 \right)=\ln 3+{{C}_{2}};$ $f\left( 1 \right)=\ln \dfrac{1}{3}+{{C}_{2}};$
$f\left( -3 \right)+f\left( 3 \right)=f\left( -1 \right)+f\left( 1 \right)=2$ $\Leftrightarrow {{C}_{1}}+{{C}_{3}}=2{{C}_{2}}=2$ $\Rightarrow \left\{ \begin{aligned}
& {{C}_{1}}+{{C}_{3}}=2 \\
& {{C}_{2}}=1 \\
\end{aligned} \right.$.
Vậy $f\left( -4 \right)+f\left( 0 \right)+f\left( 4 \right)$ $=\ln 3+{{C}_{1}}+{{C}_{2}}+\ln \dfrac{1}{3}+{{C}_{3}}$ $={{C}_{1}}+{{C}_{2}}+{{C}_{3}}=3$.
Do đó: $f\left( x \right)=\left\{ \begin{aligned}
& \ln \dfrac{x-2}{x+2}+{{C}_{1}} \text{khi} x<-2 \\
& \ln \dfrac{2-x}{x+2}+{{C}_{2}} \text{khi} -2<x<2 \\
& \ln \dfrac{x-2}{x+2}+{{C}_{3}} \text{khi} x>2 \\
\end{aligned} \right.$
$f\left( -3 \right)=\ln 5+{{C}_{1}};$ $f\left( 3 \right)=\ln \dfrac{1}{5}+{{C}_{3}};$ $f\left( 0 \right)={{C}_{2}}$ ; $f\left( -1 \right)=\ln 3+{{C}_{2}};$ $f\left( 1 \right)=\ln \dfrac{1}{3}+{{C}_{2}};$
$f\left( -3 \right)+f\left( 3 \right)=f\left( -1 \right)+f\left( 1 \right)=2$ $\Leftrightarrow {{C}_{1}}+{{C}_{3}}=2{{C}_{2}}=2$ $\Rightarrow \left\{ \begin{aligned}
& {{C}_{1}}+{{C}_{3}}=2 \\
& {{C}_{2}}=1 \\
\end{aligned} \right.$.
Vậy $f\left( -4 \right)+f\left( 0 \right)+f\left( 4 \right)$ $=\ln 3+{{C}_{1}}+{{C}_{2}}+\ln \dfrac{1}{3}+{{C}_{3}}$ $={{C}_{1}}+{{C}_{2}}+{{C}_{3}}=3$.
Đáp án D.