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Câu hỏi: Cho $\int\limits_{0}^{1}{\dfrac{dx}{\sqrt{x+2}+\sqrt{x+1}}=a\sqrt{b}-\dfrac{8}{3}\sqrt{a}+\dfrac{2}{3}\left( a,b\in {{\mathbb{R}}^{*}} \right)}$. Tính $a+2b$ ?
A. $a+2b=7.$
B. $a+2b=5.$
C. $a+2b=-1.$
D. $a+2b=8.$
A. $a+2b=7.$
B. $a+2b=5.$
C. $a+2b=-1.$
D. $a+2b=8.$
Ta có $\int\limits_{0}^{1}{\dfrac{dx}{\sqrt{x+2}+\sqrt{x+1}}=\int\limits_{0}^{1}{\left( \sqrt{x+2}-\sqrt{x+1} \right)dx=\int\limits_{0}^{1}{{{\left( x+2 \right)}^{\dfrac{1}{2}}}d\left( x+2 \right)-\int\limits_{0}^{1}{{{\left( x+1 \right)}^{\dfrac{1}{2}}}d\left( x+1 \right)}}}}$
$=\left. \dfrac{2}{3}\left( {{\left( x+2 \right)}^{\dfrac{3}{2}}}-{{\left( x+1 \right)}^{\dfrac{3}{2}}} \right) \right|_{0}^{1}=2\sqrt{3}-\dfrac{8}{3}\sqrt{2}+\dfrac{2}{3}$. Suy ra $a=2,b=3.$ Vậy $a+2b=8.$
$=\left. \dfrac{2}{3}\left( {{\left( x+2 \right)}^{\dfrac{3}{2}}}-{{\left( x+1 \right)}^{\dfrac{3}{2}}} \right) \right|_{0}^{1}=2\sqrt{3}-\dfrac{8}{3}\sqrt{2}+\dfrac{2}{3}$. Suy ra $a=2,b=3.$ Vậy $a+2b=8.$
Đáp án D.