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Câu hỏi: . Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ 1;6 \right]$ và thỏa mãn $f\left( x \right)=\dfrac{f\left( 2\sqrt{x+3}-3 \right)}{\sqrt{x+3}}+\dfrac{x}{\sqrt{x+3}}.$ Tính tích phân của $I=\int\limits_{3}^{6}{f\left( x \right)\text{d}x}$
A. $I=\dfrac{10}{3}.$
B. $I=\dfrac{20}{3}.$
C. $I=4.$
D. $I=\dfrac{10}{3}+\ln 2.$
A. $I=\dfrac{10}{3}.$
B. $I=\dfrac{20}{3}.$
C. $I=4.$
D. $I=\dfrac{10}{3}+\ln 2.$
Theo giả thiết ta có: $f\left( x \right)=\dfrac{f\left( 2\sqrt{x+3}-3 \right)}{\sqrt{x+3}}+\dfrac{x}{\sqrt{x+3}}$
Lấy tích phân hai vế cận từ 1 đến 6 ta được: $\int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{6}{\dfrac{f\left( 2\sqrt{x+3}-3 \right)}{\sqrt{x+3}}d\text{x}}+\int\limits_{1}^{6}{\dfrac{x\text{dx}}{\sqrt{x+3}}}$
$\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{6}{f\left( 2\sqrt{x+3}-3 \right)d\left( 2\sqrt{x+3}-3 \right)}+\dfrac{20}{3}$ (Casio ta được $\int\limits_{1}^{6}{\dfrac{x\text{dx}}{\sqrt{x+3}}}=\dfrac{20}{3}$ )
$\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{3}{f\left( u \right)du}+\dfrac{20}{3}\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{3}{f\left( x \right)d\text{x}}+\dfrac{20}{3}$
Do đó $I=\int\limits_{3}^{6}{f\left( x \right)dx}=\dfrac{20}{3}$.
Lấy tích phân hai vế cận từ 1 đến 6 ta được: $\int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{6}{\dfrac{f\left( 2\sqrt{x+3}-3 \right)}{\sqrt{x+3}}d\text{x}}+\int\limits_{1}^{6}{\dfrac{x\text{dx}}{\sqrt{x+3}}}$
$\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{6}{f\left( 2\sqrt{x+3}-3 \right)d\left( 2\sqrt{x+3}-3 \right)}+\dfrac{20}{3}$ (Casio ta được $\int\limits_{1}^{6}{\dfrac{x\text{dx}}{\sqrt{x+3}}}=\dfrac{20}{3}$ )
$\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{3}{f\left( u \right)du}+\dfrac{20}{3}\Leftrightarrow \int\limits_{1}^{6}{f\left( x \right)d\text{x}}=\int\limits_{1}^{3}{f\left( x \right)d\text{x}}+\dfrac{20}{3}$
Do đó $I=\int\limits_{3}^{6}{f\left( x \right)dx}=\dfrac{20}{3}$.
Đáp án B.