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Câu hỏi: Biết $\int\limits_{1}^{2}{\dfrac{\left( x-1 \right)dx}{\sqrt{2x-1}+\sqrt{x}}}=a\sqrt{3}+b\sqrt{2}+c$ với a, b, c là các số hữu tỷ. Tính $P=a+b+c.$
A. $P=1.$
B. $P=2.$
C. $P=0.$
D. $P=3.$
A. $P=1.$
B. $P=2.$
C. $P=0.$
D. $P=3.$
HD: Ta có: $I=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{2x-1}-\sqrt{x} \right)\left( x-1 \right)}{\left( 2x-1 \right)-x}dx=\int\limits_{1}^{2}{\left( \sqrt{2x-1}-\sqrt{x} \right)}}dx=\dfrac{1}{2}\sqrt{2x-1}d\left( 2x-1 \right)-\dfrac{1}{2}\int\limits_{1}^{2}{\sqrt{x}dx}$
$=\left. \left( \dfrac{1}{2}.\dfrac{2}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}-\dfrac{2}{3}\sqrt{{{x}^{3}}} \right) \right|_{1}^{2}=\sqrt{3}-\dfrac{4}{3}\sqrt{2}+\dfrac{1}{3}$
Do đó $a=1,b=\dfrac{-4}{3},c=\dfrac{1}{3}\Rightarrow a+b+c=0.$
$=\left. \left( \dfrac{1}{2}.\dfrac{2}{3}\sqrt{{{\left( 2x-1 \right)}^{3}}}-\dfrac{2}{3}\sqrt{{{x}^{3}}} \right) \right|_{1}^{2}=\sqrt{3}-\dfrac{4}{3}\sqrt{2}+\dfrac{1}{3}$
Do đó $a=1,b=\dfrac{-4}{3},c=\dfrac{1}{3}\Rightarrow a+b+c=0.$
Đáp án C.