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Câu hỏi: Biết rằng $\int\limits_{2}^{5}{\dfrac{dx}{\sqrt{4x+1}+2\sqrt{x-1}}}=\dfrac{a+b\sqrt{21}}{30},$ với $a,b\in \mathbb{Z}.$ Tính $S=a+b.$
A. $S=-1.$
B. $S=-2.$
C. $S=-54.$
D. $S=-62.$
A. $S=-1.$
B. $S=-2.$
C. $S=-54.$
D. $S=-62.$
Ta có $\left( \sqrt{4x+1}+2\sqrt{x-1} \right)\left( \sqrt{4x+1}-2\sqrt{x-1} \right)=\left( 4x+1 \right)-4\left( x-1 \right)=5$
$\Rightarrow \int\limits_{2}^{5}{\dfrac{dx}{\sqrt{4x+1}+2\sqrt{x-1}}}=\int\limits_{2}^{5}{\dfrac{\sqrt{4x+1}-2\sqrt{x-1}}{5}dx}=\dfrac{1}{5}\int\limits_{2}^{5}{\left[ {{\left( 4x+1 \right)}^{\dfrac{1}{2}}}-2{{\left( x-1 \right)}^{\dfrac{1}{2}}} \right]dx}$
$=\left. \left[ \dfrac{1}{5}.\dfrac{1}{4}.\dfrac{{{\left( 4\text{x}+1 \right)}^{\dfrac{1}{2}+1}}}{\dfrac{3}{2}}-\dfrac{2}{5}.\dfrac{{{\left( x-1 \right)}^{\dfrac{1}{2}+1}}}{\dfrac{3}{2}} \right] \right|_{2}^{5}=\left. \left[ \dfrac{1}{30}\sqrt{{{\left( 4\text{x}+1 \right)}^{3}}}-\dfrac{4}{15}\sqrt{{{\left( x-1 \right)}^{3}}} \right] \right|_{2}^{5}$
$=\left( \dfrac{21\sqrt{21}}{30}-\dfrac{32}{15} \right)-\dfrac{19}{30}=\dfrac{-83+21\sqrt{21}}{30}\Rightarrow \left\{ \begin{aligned}
& a=-83 \\
& b=21 \\
\end{aligned} \right.\Rightarrow S=-62.$
$\Rightarrow \int\limits_{2}^{5}{\dfrac{dx}{\sqrt{4x+1}+2\sqrt{x-1}}}=\int\limits_{2}^{5}{\dfrac{\sqrt{4x+1}-2\sqrt{x-1}}{5}dx}=\dfrac{1}{5}\int\limits_{2}^{5}{\left[ {{\left( 4x+1 \right)}^{\dfrac{1}{2}}}-2{{\left( x-1 \right)}^{\dfrac{1}{2}}} \right]dx}$
$=\left. \left[ \dfrac{1}{5}.\dfrac{1}{4}.\dfrac{{{\left( 4\text{x}+1 \right)}^{\dfrac{1}{2}+1}}}{\dfrac{3}{2}}-\dfrac{2}{5}.\dfrac{{{\left( x-1 \right)}^{\dfrac{1}{2}+1}}}{\dfrac{3}{2}} \right] \right|_{2}^{5}=\left. \left[ \dfrac{1}{30}\sqrt{{{\left( 4\text{x}+1 \right)}^{3}}}-\dfrac{4}{15}\sqrt{{{\left( x-1 \right)}^{3}}} \right] \right|_{2}^{5}$
$=\left( \dfrac{21\sqrt{21}}{30}-\dfrac{32}{15} \right)-\dfrac{19}{30}=\dfrac{-83+21\sqrt{21}}{30}\Rightarrow \left\{ \begin{aligned}
& a=-83 \\
& b=21 \\
\end{aligned} \right.\Rightarrow S=-62.$
Đáp án D.