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Câu hỏi: Biết $\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}}=\sqrt{a}-\sqrt{b}-\sqrt{c}$ với $a,b,c$ là các số nguyên dương. Tính $P=a+b+c.$
A. $P=44.$
B. $P=42.$
C. $P=46.$
D. $P=48.$
A. $P=44.$
B. $P=42.$
C. $P=46.$
D. $P=48.$
Đặt $I=\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x\left( x+1 \right)}\left( \sqrt{x}+\sqrt{x+1} \right)}.}}$
Đặt $t=\sqrt{x}+\sqrt{x+1}\Rightarrow dt=\dfrac{\sqrt{x}+\sqrt{x+1}}{2\sqrt{x\left( x+1 \right)}}dx\Leftrightarrow \dfrac{dx}{\sqrt{x\left( x+1 \right)}}=2\dfrac{dt}{t}.$
Khi $x=1$ thì $t=\sqrt{2}+1,$ khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}.$
$I=\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}}=2\int\limits_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}{\dfrac{dt}{{{t}^{2}}}}=-2\left. \dfrac{1}{t} \right|_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}=-2\left( \dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{2}+1} \right)$
$=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-\sqrt{4}\Rightarrow a=32,b=12,c=4.$ Vậy $P=a+b+c=48$
Đặt $t=\sqrt{x}+\sqrt{x+1}\Rightarrow dt=\dfrac{\sqrt{x}+\sqrt{x+1}}{2\sqrt{x\left( x+1 \right)}}dx\Leftrightarrow \dfrac{dx}{\sqrt{x\left( x+1 \right)}}=2\dfrac{dt}{t}.$
Khi $x=1$ thì $t=\sqrt{2}+1,$ khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}.$
$I=\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+1}+\left( x+1 \right)\sqrt{x}}}=2\int\limits_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}{\dfrac{dt}{{{t}^{2}}}}=-2\left. \dfrac{1}{t} \right|_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}}=-2\left( \dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{1}{\sqrt{2}+1} \right)$
$=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-\sqrt{4}\Rightarrow a=32,b=12,c=4.$ Vậy $P=a+b+c=48$
Đáp án D.