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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=8; \int\limits_{0}^{3}{f\left( x \right)\text{d}x}=10$. Giá trị của $\int\limits_{-1}^{1}{f\left( \left| 2x-1 \right| \right)\text{d}x}$ bằng
A. $-1.$
B. $1.$
C. $9.$
D. $-9.$
A. $-1.$
B. $1.$
C. $9.$
D. $-9.$
Đặt $t=2 x-1 \Rightarrow \int_{-1}^{1} f(|2 x-1|) \mathrm{d} x=\int_{-3}^{1} f(|t|) \cdot \dfrac{\mathrm{d} t}{2}=\dfrac{1}{2} \int_{-1}^{3} f(|t|) \mathrm{d} t=\dfrac{1}{2}\left(\int_{-3}^{0} f(-t) \mathrm{d} t+\int_{0}^{1} f(t) \mathrm{d} t\right)$
$=\dfrac{1}{2}\left(\int_{0}^{3} f(t) \mathrm{d} t+\int_{0}^{1} f(t) \mathrm{d} t\right)=\dfrac{1}{2}(10+8)=9 .$
$=\dfrac{1}{2}\left(\int_{0}^{3} f(t) \mathrm{d} t+\int_{0}^{1} f(t) \mathrm{d} t\right)=\dfrac{1}{2}(10+8)=9 .$
Đáp án C.