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Câu hỏi: Biết $\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+2}+\left( x+2 \right)\sqrt{x}}=\sqrt{a}+\sqrt{b}-c}$ với a, b, c là các số nguyên dương. Tính $P=a+b+c$.
A. P = 2
B. P = 8
C. P = 46
D. P = 22
A. P = 2
B. P = 8
C. P = 46
D. P = 22
Ta có
$\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+2}+\left( x+2 \right)\sqrt{x}}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x}\sqrt{x+2}\left( \sqrt{x+2}+\sqrt{x} \right)}=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+2}-\sqrt{x} \right)}{2\sqrt{x}\sqrt{x+2}}dx}}$
$=\int\limits_{1}^{2}{\left( \dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x+2}} \right)dx}=\left( \sqrt{x}-\sqrt{x+2} \right)\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.=\sqrt{2}+\sqrt{3}-3$
Vậy $a=2; b=3; c=3$ (hoặc $a=2; b=3; c=3$ ) nên $P=a+b+c=8$
$\int\limits_{1}^{2}{\dfrac{dx}{x\sqrt{x+2}+\left( x+2 \right)\sqrt{x}}}=\int\limits_{1}^{2}{\dfrac{dx}{\sqrt{x}\sqrt{x+2}\left( \sqrt{x+2}+\sqrt{x} \right)}=\int\limits_{1}^{2}{\dfrac{\left( \sqrt{x+2}-\sqrt{x} \right)}{2\sqrt{x}\sqrt{x+2}}dx}}$
$=\int\limits_{1}^{2}{\left( \dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x+2}} \right)dx}=\left( \sqrt{x}-\sqrt{x+2} \right)\left| \begin{aligned}
& ^{2} \\
& _{1} \\
\end{aligned} \right.=\sqrt{2}+\sqrt{3}-3$
Vậy $a=2; b=3; c=3$ (hoặc $a=2; b=3; c=3$ ) nên $P=a+b+c=8$
Đáp án B.