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Câu hỏi: Biết $\int_1^2 \dfrac{\mathrm{d} x}{x \sqrt{x+1}+(x+1) \sqrt{x}}=\sqrt{a}-\sqrt{b}-\sqrt{c}$ 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 $I=\int_1^2 \dfrac{\mathrm{d} x}{x \sqrt{x+1}+(x+1) \sqrt{x}}=\int_1^2 \dfrac{\mathrm{d} x}{\sqrt{x(x+1)(\sqrt{x}+\sqrt{x+1})}}$.
Đặt $t=\sqrt{x}+\sqrt{x+1} \Rightarrow \mathrm{d} t=\dfrac{\sqrt{x+1}+\sqrt{x}}{2 \sqrt{x(x+1)}} \mathrm{d} x \Leftrightarrow \dfrac{\mathrm{d} x}{\sqrt{x(x+1)}}=2 \dfrac{\mathrm{d} t}{t}$.
Khi $x=1$ thì $t=\sqrt{2}+1$, khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}$.
$
\begin{aligned}
& I=\int_1^2 \dfrac{\mathrm{d} x}{\sqrt{x(x+1)}(\sqrt{x}+\sqrt{x+1})}=2 \int_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}} \dfrac{\mathrm{d} t}{t^2}=-\left.2 \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 \\
& \text { Vậy } P=a+b+c=48
\end{aligned}
$
Đặt $t=\sqrt{x}+\sqrt{x+1} \Rightarrow \mathrm{d} t=\dfrac{\sqrt{x+1}+\sqrt{x}}{2 \sqrt{x(x+1)}} \mathrm{d} x \Leftrightarrow \dfrac{\mathrm{d} x}{\sqrt{x(x+1)}}=2 \dfrac{\mathrm{d} t}{t}$.
Khi $x=1$ thì $t=\sqrt{2}+1$, khi $x=2$ thì $t=\sqrt{3}+\sqrt{2}$.
$
\begin{aligned}
& I=\int_1^2 \dfrac{\mathrm{d} x}{\sqrt{x(x+1)}(\sqrt{x}+\sqrt{x+1})}=2 \int_{\sqrt{2}+1}^{\sqrt{3}+\sqrt{2}} \dfrac{\mathrm{d} t}{t^2}=-\left.2 \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 \\
& \text { Vậy } P=a+b+c=48
\end{aligned}
$
Đáp án D.